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</div><h2>SL Paper 1</h2><div class="specification">
<p>Consider the simultaneous linear equations</p>
<p style="padding-left: 180px;">\(x + z = - 1\)<br>\(3x + y + 2z = 1\)<br>\(2x + ay - z = b\)</p>
<p>where \(a\) and \(b\) are constants.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Using row reduction, find the solutions in terms of \(a\) and \(b\) when \(a\) ≠ 3 .</p>
<div class="marks">[8]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Explain why the equations have no unique solution when \(a\) = 3.</p>
<div class="marks">[1]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find all the solutions to the equations when \(a\) = 3, \(b\) = 10 in the form <em><strong>r</strong></em> = <em><strong>s</strong></em> + \(\lambda \)<em><strong>t</strong></em>.</p>
<div class="marks">[4]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\(\left({\begin{array}{*{20}{c}}
1 \\
3 \\
2
\end{array}\,\,
\begin{array}{*{20}{c}}
0 \\
1 \\
a
\end{array}\,\,
\begin{array}{*{20}{c}}
1 \\
2 \\
{-1}
\end{array}\,\,
\begin{array}{*{20}{c}}
{-1} \\
1 \\
b
\end{array}}
\right)
\Rightarrow
\left(
{\begin{array}{*{20}{c}}
1 \\
{3a - 2} \\
2
\end{array}\,\,
\begin{array}{*{20}{c}}
0 \\
0 \\
a
\end{array}\,\,
\begin{array}{*{20}{c}}
1 \\
{2a+1} \\
{-1}
\end{array}\,\,
\begin{array}{*{20}{c}}
{-1} \\
{a-b} \\
b
\end{array}}
\right)\) or equivalent <em><strong>M1A1</strong></em></p>
<p>\(\left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> {3a - 2} \\ <br> 2 <br>\end{array}\,\,\begin{array}{*{20}{c}}<br> 0 \\ <br> 0 \\ <br> a <br>\end{array}\,\,\begin{array}{*{20}{c}}<br> {a - 3} \\ <br> {2a + 1} \\ <br> { - 1} <br>\end{array}\,\,\begin{array}{*{20}{c}}<br> { - 4a + 2 + b} \\ <br> {a - b} \\ <br> b <br>\end{array}} \right)\)</p>
<p>\(z = \frac{{ - 4a + b + 2}}{{a - 3}}\) <em><strong>M1A1</strong></em></p>
<p>\(x = - 1 - z\) <em><strong>M1</strong></em></p>
<p>\(x = - 1 - \left( {\frac{{ - 4a + b + 2}}{{a - 3}}} \right)\)</p>
<p>\(x = \frac{{ - a + 3 + 4a - b - 2}}{{a - 3}}\)</p>
<p>\(x = \frac{{3a - b + 1}}{{a - 3}}\) <em><strong>A1</strong></em></p>
<p>\(y = 1 - 3x - 2z\) <em><strong>M1</strong></em></p>
<p>\(y = 1 - 3\left( {\frac{{3a - b + 1}}{{a - 3}}} \right) - 2\left( {\frac{{ - 4a + b + 2}}{{a - 3}}} \right)\)</p>
<p>\( = \frac{{a - 3 - 9a + 3b - 3 + 8a - 2b - 4}}{{a - 3}}\)</p>
<p>\( = \frac{{b - 10}}{{a - 3}}\) <em><strong>A1</strong></em></p>
<p><em><strong>[8 marks]</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>when \(a\) = 3 the denominator of \(x\), \(y\) and \(z\) = 0 <em><strong>R1</strong></em></p>
<p><strong>Note:</strong> Accept any valid reason.</p>
<p>hence no unique solutions <em><strong>AG</strong></em></p>
<p><em><strong>[1 mark]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>For example let \(z = \lambda \) <em><strong>(M1)</strong></em></p>
<p>\(x = - 1 - \lambda \) <em><strong>(A1)</strong></em></p>
<p>\(y = 1 - 3\left( { - 1 - \lambda } \right) - 2\lambda \)</p>
<p>\(y = 4 + \lambda \) <em><strong>(A1)</strong></em></p>
<p><em>r</em> = \(\left( \begin{gathered}<br> - 1 \hfill \\<br> 4 \hfill \\<br> 0 \hfill \\ <br>\end{gathered} \right) + \lambda \left( \begin{gathered}<br> - 1 \hfill \\<br> 1 \hfill \\<br> 1 \hfill \\ <br>\end{gathered} \right)\) <em><strong>A1</strong></em></p>
<p><em><strong>[4 marks]</strong></em></p>
<p><strong>Note</strong>: Accept answers which let \(x = \lambda \) or \(y = \lambda \).</p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p>Consider the matrix <em><strong>M</strong></em> = \(\left[ {\begin{array}{*{20}{c}}<br> 2 \\ <br> { - 1} <br>\end{array}\,\,\,\begin{array}{*{20}{c}}<br> { - 4} \\ <br> { - 1} <br>\end{array}} \right]\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that the linear transformation represented by <em><strong>M</strong></em> transforms any point on the line \(y = x\) to a point on the same line.</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Explain what happens to points on the line \(4y + x = 0\) when they are transformed by <strong><em>M</em></strong>.</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>State the two eigenvalues of <strong><em>M</em></strong>.</p>
<div class="marks">[2]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>State two eigenvectors of <em><strong>M</strong></em> which correspond to the two eigenvalues.</p>
<div class="marks">[2]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\(\left( {\begin{array}{*{20}{c}}<br> 2 \\ <br> { - 1} <br>\end{array}\,\,\,\begin{array}{*{20}{c}}<br> { - 4} \\ <br> { - 1} <br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br> k \\ <br> k <br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br> { - 2k} \\ <br> { - 2k} <br>\end{array}} \right)\left( { = - 2\left( {\begin{array}{*{20}{c}}<br> k \\ <br> k <br>\end{array}} \right)} \right)\) <em><strong>M1A1</strong></em></p>
<p>hence still on the line \(y = x\) <em><strong>AG</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>consider \(\left( {\begin{array}{*{20}{c}}<br> 2 \\ <br> { - 1} <br>\end{array}\,\,\,\begin{array}{*{20}{c}}<br> { - 4} \\ <br> { - 1} <br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br> {4k} \\ <br> { - k} <br>\end{array}} \right)\) <em><strong>M1</strong></em></p>
<p>\( = \left( {\begin{array}{*{20}{c}}<br> {12k} \\ <br> { - 3k} <br>\end{array}} \right)\left( { = 3\left( {\begin{array}{*{20}{c}}<br> {4k} \\ <br> { - k} <br>\end{array}} \right)} \right)\) <em><strong>A1</strong></em></p>
<p>hence the line is invariant <em><strong>A1</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>hence the eigenvalues are −2 and 3 <em><strong>A</strong><strong>1A1</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> 1 <br>\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}<br> 4 \\ <br> { - 1} <br>\end{array}} \right)\) or equivalent <em><strong>A</strong><strong>1A1</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">A matrix <span class="s1"><strong><em>M </em></strong></span>is called idempotent if <span class="s1"><strong><em>M</em></strong></span><span class="s2">\(^2 = \) </span><span class="s1"><strong><em>M</em></strong></span>.</p>
</div>
<div class="specification">
<p class="p1">The idempotent matrix <span class="s1"><strong><em>N </em></strong></span>has the form</p>
<p class="p1" style="text-align: center;"><strong><em>N</em></strong> \( = \left( {\begin{array}{*{20}{c}} a&{ - 2a} \\ a&{ - 2a} \end{array}} \right)\)</p>
<p class="p1" style="text-align: left;">where \(a \ne 0\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Explain why <span class="s1"><strong><em>M </em></strong></span>is a square matrix.</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Find the set of possible values of <span class="s1">det(<strong><em>M</em></strong>).</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Find the value of \(a\).</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Find the eigenvalues of <span class="s1"><strong><em>N</em></strong></span><span class="s2">.</span></p>
<p class="p1">(iii) <span class="Apple-converted-space"> </span>Find corresponding eigenvectors.</p>
<div class="marks">[12]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>M</em></strong></span><span class="s2">\(^2 = \) </span><span class="s1"><strong><em>MM </em></strong></span>only exists if the number of columns of <span class="s3"><strong><em>M </em></strong></span>equals the number of rows of <span class="s3"><strong><em>M <span class="Apple-converted-space"> </span></em></strong></span><strong><em>R1</em></strong></p>
<p class="p1">hence <span class="s1"><strong><em>M </em></strong></span>is square <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>apply the determinant function to both sides <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2">\(\det (\)<strong><em>M</em></strong>\(^2) = \det (\)<strong><em>M</em></strong>\()\)</p>
<p class="p2">use the multiplicative property of the determinant</p>
<p class="p2">\(\det (\)<strong><em>M</em></strong>\(^2) = \det (\)<strong><em>M</em></strong>\(){\text{ }}\det (\)<strong><em>M</em></strong>\() = \det (\)<strong><em>M</em></strong>\()\) <span class="Apple-converted-space"> </span><span class="s4"><strong><em>(M1)</em></strong></span></p>
<p class="p2">hence \(\det (\)<strong><em>M</em></strong>\() = 0\) or 1 <span class="Apple-converted-space"> </span><span class="s4"><strong><em>A1</em></strong></span></p>
<p class="p1"><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>attempt to calculate <span class="s1"><strong><em>N</em></strong></span>\(^2\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">obtain \(\left( {\begin{array}{*{20}{c}} { - {a^2}}&{2{a^2}} \\ { - {a^2}}&{2{a^2}} \end{array}} \right)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">equating to <span class="s1"><strong><em>N <span class="Apple-converted-space"> </span></em></strong></span><strong><em>M1</em></strong></p>
<p class="p1">to obtain \(a = - 1\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p2"> </p>
<p class="p2">(ii) <span class="Apple-converted-space"> </span><strong><em>N</em></strong> \( = \left( {\begin{array}{*{20}{c}} { - 1}&2 \\ { - 1}&2 \end{array}} \right)\)</p>
<p class="p2"><strong><em>N</em></strong> \( - \lambda \)<strong><em>I</em></strong> \( = \left( {\begin{array}{*{20}{c}} { - 1 - \lambda }&2 \\ { - 1}&{2 - \lambda } \end{array}} \right)\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>M1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\(( - 1 - \lambda )(2 - \lambda ) + 2 = 0\) </span><span class="s2"><strong><em>(A1)</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\({\lambda ^2} - \lambda = 0\) </span><span class="s2"><strong><em>(A1)</em></strong></span></p>
<p class="p2">\(\lambda \) is 1 or 0 <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p1"> </p>
<p class="p1">(iii) <span class="Apple-converted-space"> </span>let \(\lambda = 1\)</p>
<p class="p2">to obtain \(\left( {\begin{array}{*{20}{c}} { - 1}&2 \\ { - 1}&2 \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right){\text{ or }}\left( {\begin{array}{*{20}{c}} { - 2}&2 \\ { - 1}&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right)\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>M1</em></strong></span></p>
<p class="p1">hence eigenvector is \(\left( {\begin{array}{*{20}{c}} x \\ x \end{array}} \right)\)<span class="s1"> <span class="Apple-converted-space"> </span></span><strong><em>A1</em></strong></p>
<p class="p1">let \(\lambda = 0\)</p>
<p class="p2">to obtain \(\left( {\begin{array}{*{20}{c}} { - 1}&2 \\ { - 1}&2 \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right)\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>M1</em></strong></span></p>
<p class="p1">hence eigenvector is \(\left( {\begin{array}{*{20}{c}} {2y} \\ y \end{array}} \right)\)<span class="s1"> <span class="Apple-converted-space"> </span></span><strong><em>A1</em></strong></p>
<p class="p3"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Accept specific eigenvectors.</p>
<p class="p3"> </p>
<p class="p1"><strong><em>[12 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">This was a more successful question for many candidates with a number of fully correct solutions being seen and a significant number of partially correct answers. Most candidates understood what was required from part (a)(i), but part (ii) often resulted in unnecessarily complex algebra which they were unable to manipulate. Part (b) resulted in many wholly successful answers, provided candidates realised the need for care in terms of the manipulation.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">This was a more successful question for many candidates with a number of fully correct solutions being seen and a significant number of partially correct answers. Most candidates understood what was required from part (a)(i), but part (ii) often resulted in unnecessarily complex algebra which they were unable to manipulate. Part (b) resulted in many wholly successful answers, provided candidates realised the need for care in terms of the manipulation.</p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p>Let <em><strong>A</strong></em><sup>2</sup> = 2<em><strong>A</strong></em> + <em><strong>I</strong></em> where <em><strong>A</strong></em> is a 2 × 2 matrix.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that <em><strong>A</strong></em><sup>4</sup> = 12<em><strong>A</strong></em> + 5<em><strong>I</strong></em>.</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Let <em><strong>B</strong></em> = \(\left[ {\begin{array}{*{20}{c}}<br> 4&2 \\ <br> 1&{ - 3} <br>\end{array}} \right]\).</p>
<p>Given that <em><strong>B</strong></em><sup>2</sup> – <em><strong>B</strong></em> – 4<em><strong>I</strong></em> = \(\left[ {\begin{array}{*{20}{c}}<br> k&0 \\ <br> 0&k <br>\end{array}} \right]\), find the value of \(k\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1</strong><br><em><strong>A</strong></em><sup>4</sup> = 4<em><strong>A</strong></em><sup>2</sup> + 4A<em><strong>I</strong></em> + <em><strong>I</strong></em><sup>2</sup> or equivalent <em><strong>M1A1</strong></em><br>= 4(2<em><strong>A</strong></em> + <em><strong>I</strong></em>) + 4<em><strong>A</strong></em> + <em><strong>I</strong></em> <em><strong>A1</strong></em><br>= 8<em><strong>A</strong></em> + 4I + 4<em><strong>A</strong></em> + <em><strong>I</strong></em><br>= 12<em><strong>A</strong></em> + 5<em><strong>I</strong></em> <em><strong>AG</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<p><strong>METHOD 2</strong><br><em><strong>A</strong></em><sup>3</sup> = <em><strong>A</strong></em>(2<em><strong>A</strong></em> + <em><strong>I</strong></em>) = 2<em><strong>A</strong></em><sup>2</sup> + <em><strong>A</strong><strong>I</strong></em> = 2(2<em><strong>A</strong></em> + <em><strong>I</strong></em>) + <em><strong>A</strong></em>(= 5<em><strong>A </strong></em>+ 2<em><strong>I</strong></em>) <em><strong>M1A1</strong></em><br><em><strong>A</strong></em><sup>4</sup> = <em><strong>A</strong></em>(5<em><strong>A </strong></em>+ 2<em><strong>I</strong></em>) <em><strong>A1</strong></em><br>= 5<em><strong>A</strong></em><sup>2</sup> + 2<em><strong>A</strong></em> = 5(2<em><strong>A </strong></em>+ <em><strong>I</strong></em>) + 2<em><strong>A</strong></em><br>= 12<em><strong>A</strong></em> + 5<em><strong>I</strong></em> <em><strong>AG</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><em><strong>B</strong></em><sup>2</sup> = \(\left[ {\begin{array}{*{20}{c}}<br> {18}&2 \\ <br> 1&{11} <br>\end{array}} \right]\) <em><strong> (A1)</strong></em></p>
<p>\(\left[ {\begin{array}{*{20}{c}}<br> {18}&2 \\ <br> 1&{11} <br>\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}<br> 4&2 \\ <br> 1&{ - 3} <br>\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}<br> 4&0 \\ <br> 0&4 <br>\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}<br> {10}&0 \\ <br> 0&{10} <br>\end{array}} \right]\) <em><strong>(A1)</strong></em></p>
<p>\( \Rightarrow k = 10\) <em><strong>A1</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p>Consider the system of equations</p>
<p>\[\left[ {\begin{array}{*{20}{l}} 1&2&1&3 \\ 2&1&3&1 \\ 5&1&8&0 \\ 3&3&4&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}} \\ {{x_2}} \\ {{x_3}} \\ {{x_4}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2 \\ 3 \\ \lambda \\ \mu \end{array}} \right]\]</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Determine the value of \(\lambda \) and the value of \(\mu \) for which the equations are consistent.</p>
<div class="marks">[5]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>For these values of \(\lambda \) and \(\mu \), solve the equations.</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>State the rank of the matrix of coefficients, justifying your answer.</p>
<div class="marks">[2]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>using row operations on \(4 \times 5\) matrix, <strong><em>M1</em></strong></p>
<p>\(\left[ {\begin{array}{*{20}{c}} 1&2&1&3 \\ 0&{ - 3}&1&{ - 5} \\ 0&{ - 9}&3&{ - 15} \\ 0&{ - 3}&1&{ - 5} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ {\lambda - 10} \\ {\mu - 6} \end{array}} \right]\,\,\,\begin{array}{*{20}{c}} {} \\ {{\text{row}}2 - 2 \times {\text{row1}}} \\ {{\text{row}}3 - 5 \times {\text{row1}}} \\ {{\text{row}}4 - 3 \times {\text{row1}}} \end{array}\) <strong><em>A2</em></strong></p>
<p>or any alternative correct row reductions</p>
<p> </p>
<p><strong>Note:</strong> Award <strong><em>A1 </em></strong>for two correct row reductions.</p>
<p> </p>
<p>\(\lambda = 7\) <strong><em>A1</em></strong></p>
<p>\(\mu = 5\) <strong><em>A1</em></strong></p>
<p><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>let \({x_3} = \alpha ,{\text{ }}{x_4} = \beta \) <strong><em>M1</em></strong></p>
<p>\({x_2} = \frac{{1 + \alpha - 5\beta }}{3}\) <strong><em>A1</em></strong></p>
<p>\({x_1} = \frac{{4 - 5\alpha + \beta }}{3}\) <strong><em>A1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> Alternative solutions are available.</p>
<p> </p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>the rank is 2 <strong><em>A1</em></strong></p>
<p>because the matrix has 2 independent rows or a correct comment based on the use of rref <strong><em>R1</em></strong></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p>The non-zero vectors <strong><em>v</em></strong><sub>1</sub>, <strong><em>v</em></strong><sub>2</sub>, <strong><em>v</em></strong><sub>3</sub> form an orthogonal set of vectors in \({\mathbb{R}^3}\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>By considering \({\alpha _1}\)<strong><em>v</em></strong>\(_1 + {\alpha _2}\)<strong><em>v</em></strong>\(_2 + {\alpha _3}\)<strong><em>v</em></strong>\(_3 = 0\), show that <strong><em>v</em></strong>\(_1\), <strong><em>v</em></strong>\(_2\), <strong><em>v</em></strong>\(_3\) are linearly independent.</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Explain briefly why <strong><em>v</em></strong>\(_1\), <strong><em>v</em></strong>\(_2\), <strong><em>v</em></strong>\(_3\) form a basis for vectors in \({\mathbb{R}^3}\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that the vectors</p>
<p>\[\left[ {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right];{\text{ }}\left[ {\begin{array}{*{20}{c}} { - 1} \\ 1 \\ 1 \end{array}} \right];{\text{ }}\left[ {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right]\]</p>
<p>form an orthogonal basis.</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Express the vector</p>
<p>\[\left[ {\begin{array}{*{20}{c}} 2 \\ 8 \\ 0 \end{array}} \right]\]</p>
<p>as a linear combination of these vectors.</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.ii.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>let \({\alpha _1}\)<strong><em>v</em></strong>\(_1 + {\alpha _2}\)<strong><em>v</em></strong>\(_2 + {\alpha _3}\)<strong><em>v</em></strong>\(_3 = 0\)</p>
<p>take the dot product with <strong><em>v</em></strong>\(_1\) <strong><em>M1</em></strong></p>
<p>\({\alpha _1}\)<strong><em>v</em></strong>\(_1\) \( \bullet \) <strong><em>v</em></strong>\(_1 + {\alpha _2}\)<strong><em>v</em></strong>\(_2\) \( \bullet \) <strong><em>v</em></strong>\(_1 + {\alpha _3}\)<strong><em>v</em></strong>\(_3\) \( \bullet \) <strong><em>v</em></strong>\(_1 = 0\) <strong><em>A1</em></strong></p>
<p>because the vectors are orthogonal, <strong><em>v</em></strong>\(_2\) \( \bullet \) <strong><em>v</em></strong>\(_1\) = <strong><em>v</em></strong>\(_3\) \( \bullet \) <strong><em>v</em></strong>\(_1\) = 0 <strong><em>R1</em></strong></p>
<p>and since <strong><em>v</em></strong>\(_1\) \( \bullet \) <strong><em>v</em></strong>\(_1\) > 0 it follows that \({\alpha _1} = 0\) and similarly, \({\alpha _2} = {\alpha _3} = 0\) <strong><em>R1</em></strong></p>
<p>so \({\alpha _1}\)<strong><em>v</em></strong>\(_1 + {\alpha _2}\)<strong><em>v</em></strong>\(_2 + {\alpha _3}\)<strong><em>v</em></strong>\(_3 = 0 \Rightarrow {\alpha _1} = {\alpha _2} = {\alpha _3} = 0\) therefore <strong><em>v</em></strong>\(_1\), <strong><em>v</em></strong>\(_2\), <strong><em>v</em></strong>\(_3\), are linearly independent <strong><em>R1AG</em></strong></p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>the three vectors form a basis for \({\mathbb{R}^3}\) because they are (linearly) independent <strong><em>R1</em></strong></p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\left[ {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right] \bullet \left[ {\begin{array}{*{20}{c}} { - 1} \\ 1 \\ 1 \end{array}} \right] = 0;{\text{ }}\left[ {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right] \bullet \left[ {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right] = 0;{\text{ }}\left[ {\begin{array}{*{20}{c}} { - 1} \\ 1 \\ 1 \end{array}} \right] \bullet \left[ {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right] = 0\) <strong><em>M1A1</em></strong></p>
<p>therefore the vectors form an orthogonal basis <strong><em>AG</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>let \(\left[ {\begin{array}{*{20}{c}} 2 \\ 8 \\ 0 \end{array}} \right] = \lambda \left[ {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right] + \mu \left[ {\begin{array}{*{20}{c}} { - 1} \\ 1 \\ 1 \end{array}} \right] + v\left[ {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right]\) <strong><em>M1</em></strong></p>
<p>\(\lambda - \mu + v = 2\)</p>
<p>\(\mu + 2v = 8\)</p>
<p>\(\lambda = \mu - v = 0\) <strong><em>A1</em></strong></p>
<p>the solution is</p>
<p>\(\left[ {\begin{array}{*{20}{c}} \lambda \\ \mu \\ v \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\ 2 \\ 3 \end{array}} \right]\,\,\,\,\,\left( {\left[ {\begin{array}{*{20}{c}} 2 \\ 8 \\ 0 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right] + 2\left[ {\begin{array}{*{20}{c}} { - 1} \\ 1 \\ 1 \end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right]} \right)\) <strong><em>A1</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.ii.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.ii.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">In this question, \(x\)<span class="s1">, </span>\(y\) and \(z\) denote the coordinates of a point in three-dimensional Euclidean space with respect to fixed rectangular axes with origin O. The vector space of position vectors relative to <span class="s1">O </span>is denoted by \({\mathbb{R}^3}\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Explain why the set of position vectors of points whose coordinates satisfy \(x - y - z = 1\) <span class="s1">does not form a vector subspace of \({\mathbb{R}^3}\).</span></p>
<div class="marks">[1]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Show that the set of position vectors of points whose coordinates satisfy \(x - y - z = 0\) forms a vector subspace, \(V\), of \({\mathbb{R}^3}\).</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Determine an orthogonal basis for \(V\) <span class="s1">of which one member is \(\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right)\).</span></p>
<p class="p2">(iii) <span class="Apple-converted-space"> </span>Augment this basis with an orthogonal vector to form a basis for \({\mathbb{R}^3}\).</p>
<p class="p1">(iv) <span class="Apple-converted-space"> </span>Express the position vector of the point with coordinates \((4,{\text{ }}0,{\text{ }} - 2)\) as a linear combination of these basis vectors.</p>
<div class="marks">[13]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">Accept any valid reasoning:</p>
<p class="p1"><strong>Example 1:</strong></p>
<p class="p2">\((1,{\text{ }}0,{\text{ }}0)\) lies on the plane, however linear combinations of this do not (for example \((2,{\text{ }}0,{\text{ }}0)\)<span class="s1">) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></span></p>
<p class="p1">hence the position vectors of the points on the plane do not form a vector space <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1"><strong>Example 2:</strong></p>
<p class="p1">the given plane does not pass through the origin (or the zero vector is not the position vector of any point on the plane) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">hence the position vectors of the points on the plane do not form a vector space <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1"><strong><em>[1 mark]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>(the set of position vectors is non-empty)</p>
<p class="p1">let \(x\)\(_1 = \left( {\begin{array}{*{20}{c}} {{x_1}} \\ {{y_1}} \\ {{z_1}} \end{array}} \right)\) be the position vector of a point on the plane and \(a \in \mathbb{R}\)</p>
<p class="p1">then the coordinates of the position vector of \(ax\)<span class="s1"><em> </em></span>satisfy the equation for the plane because \(a{x_1} - a{y_1} - a{z_1} = a({x_1} - {y_1} - {z_1}) = 0\) <strong><em>M1A1</em></strong></p>
<p class="p1">let \(x\)\(_2 = \left( {\begin{array}{*{20}{c}} {{x_2}} \\ {{y_2}} \\ {{x_2}} \end{array}} \right)\) be the position vector of another point on the plane</p>
<p class="p1">consider \(x\)\(_3 = \) \(x\)\(_1 + \) \(x\)\(_2\)</p>
<p class="p1">then the coordinates of \(x\)\(_3 = \left( {\begin{array}{*{20}{l}} {{x_3} = {x_1} + {x_2}} \\ {{y_1} = {y_1} + {y_2}} \\ {{z_3} = {z_1} + {z_2}} \end{array}} \right)\) satisfy <strong><em>M1</em></strong></p>
<p class="p1">\({x_3} - {y_3} - {z_3} = ({x_1} + {x_2}) - ({y_1} + {y_2}) - ({z_1} + {z_2})\)</p>
<p class="p1"><span class="Apple-converted-space">\( = 0\) </span><strong><em>A1</em></strong></p>
<p class="p1">subspace conditions established <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p2"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>The above conditions may be combined in one calculation.</p>
<p class="p2"> </p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>if \(\left( {\begin{array}{*{20}{c}} a \\ b \\ c \end{array}} \right)\) is the position vector of a second point on the plane orthogonal to the given vector, then <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p3"><span class="s2">\(a - b - c = 0\) </span>and \(a + 2b - c = 0\) <span class="Apple-converted-space"> </span><span class="s3"><strong><em>(A1)(A1)</em></strong></span></p>
<p class="p1"><span class="s1">for example \(\left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right)\) </span>completes the basis <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">(iii) <span class="Apple-converted-space"> </span>the basis for \(({\mathbb{R}^3})\) can be augmented to an orthogonal basis for \({\mathbb{R}^3}\) <span class="s1">by adjoining \(\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right)\) <span class="Apple-converted-space"> </span></span><strong><em>(M1)</em></strong></p>
<p class="p3"><span class="Apple-converted-space">\( = \left( {\begin{array}{*{20}{c}} 2 \\ { - 2} \\ { - 2} \end{array}} \right)\) </span><span class="s3"><strong><em>A1</em></strong></span></p>
<p class="p3">(iv) <span class="Apple-converted-space"> </span>attempt to solve \(\alpha = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right) + \beta \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right) + \gamma \left( {\begin{array}{*{20}{c}} 2 \\ { - 2} \\ { - 2} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 4 \\ 0 \\ { - 2} \end{array}} \right)\) <span class="Apple-converted-space"> </span><span class="s3"><strong><em>M1</em></strong></span></p>
<p class="p1">obtain \(\alpha = \beta = \gamma = 1\) <span class="Apple-converted-space"> </span><strong><em>A2</em></strong></p>
<p class="p1"><strong><em>[13 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">Again this was found difficult by many candidates and resulted in no attempt being made. For those who were able to start, parts (a) and (b)(i) showed a reasonable degree of understanding. After that it was only a significant minority of candidates who were able to proceed successfully with many ignoring or not realising the significance of the word “orthogonal”.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Again this was found difficult by many candidates and resulted in no attempt being made. For those who were able to start, parts (a) and (b)(i) showed a reasonable degree of understanding. After that it was only a significant minority of candidates who were able to proceed successfully with many ignoring or not realising the significance of the word “orthogonal”.</p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">The matrix <strong><em>A</em></strong> is given by <strong><em>A</em></strong> = \(\left( {\begin{array}{*{20}{c}}1&2&1\\1&1&2\\2&3&1\end{array}} \right)\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Given that <strong><em>A</em></strong>\(^3\) can be expressed in the form <strong><em>A</em></strong>\(^3 = a\)<strong><em>A</em></strong>\(^2 = b\)<strong><em>A</em></strong> \( + c\)<strong><em>I</em></strong>, determine the values of the constants \(a\), \(b\), \(c\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(b) (i) Hence express <strong><em>A</em></strong>\(^{ - 1}\) in the form <strong><em>A</em></strong>\(^{ - 1} = d\)<strong><em>A</em></strong>\(^2 = e\)<strong><em>A</em></strong> \( + f\)<strong><em>I</em></strong> where \(d,{\text{ }}e,{\text{ }}f \in \mathbb{Q}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Use this result to determine <strong><em>A</em></strong>\(^{ - 1}\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) successive powers of <strong><em>A</em></strong> are given by</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>A</em></strong>\(^2 = \) \(\left( {\begin{array}{*{20}{c}}5&7&6\\6&9&5\\7&{10}&9\end{array}} \right)\) <strong><em>(M1)A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>A</em></strong>\(^3 = \) \(\left( {\begin{array}{*{20}{c}}{24}&{35}&{25}\\{25}&{36}&{29}\\{35}&{51}&{36}\end{array}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">it follows, considering elements in the first rows, that</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(5a + b + c = 24\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(7a + 2b = 35\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(6a + b = 25\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">solving, <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\((a,{\text{ }}b,{\text{ }}c) = (3,{\text{ }}7,{\text{ }}2)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Accept any other three correct equations.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Accept the use of the Cayley–Hamilton Theorem.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[7 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) (i) it has been shown that</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>A</em></strong>\(^3 = 3\)<strong><em>A</em></strong>\(^2 + 7\)<strong><em>A</em></strong>\( + 2\)<strong><em>I</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">multiplying by <strong><em>A</em></strong>\(^{ - 1}\), <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>A</em></strong>\(^2 = 3\)<strong><em>A</em></strong>\( + 7\)<strong><em>I</em></strong>\( + 2\)<strong><em>A</em></strong>\(^{ - 1}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">whence</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>A</em></strong>\(^{ - 1} = 0.5\)<strong><em>A</em></strong>\(^2 - 1.5\)<strong><em>A</em></strong> \( - 3.5\)<strong><em>I</em></strong> <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) substituting powers of <strong><em>A</em></strong>,</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>A</em></strong>\(^{ - 1} = 0.5\)\(\left( {\begin{array}{*{20}{c}}5&7&6\\6&9&5\\7&{10}&9\end{array}} \right) - 1.5\left( {\begin{array}{*{20}{c}}1&2&1\\1&1&2\\2&3&1\end{array}} \right) - 3.5\left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">=\(\left( {\begin{array}{*{20}{c}}{ - 2.5}&{0.5}&{1.5}\\{1.5}&{ - 0.5}&{ - 0.5}\\{0.5}&{0.5}&{ - 0.5}\end{array}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Follow through their equation in (b)(i).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Line (ii) of (ii) must be seen.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="specification">
<p class="p1">A transformation \(T\) is a linear mapping from \({\mathbb{R}^3}\) to \({\mathbb{R}^4}\), represented by the matrix</p>
<p class="p1">\[M = \left( {\begin{array}{*{20}{c}} 1&2&1 \\ 2&7&5 \\ { - 3}&1&4 \\ 1&5&4 \end{array}} \right)\]</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Find the row rank of \(M\).</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Hence or otherwise find the kernel of \(T\).</p>
<div class="marks">[8]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>State the column rank of \(M\).</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Find the basis for the range of this transformation.</p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) row reduction gives \(\left( {\begin{array}{*{20}{c}} 1&2&1 \\ 0&3&3 \\ 0&7&7 \\ 0&3&3 \end{array}} \right) \to \left( {\begin{array}{*{20}{c}} 1&2&1 \\ 0&3&3 \\ 0&0&0 \\ 0&0&0 \end{array}} \right)\left( { \to \left( {\begin{array}{*{20}{c}} 1&0&{ - 1} \\ 0&1&1 \\ 0&0&0 \\ 0&0&0 \end{array}} \right)} \right)\) <strong><em>(M1)A1</em></strong></p>
<p class="p1">hence row rank is \(2\) <strong><em>A1</em></strong></p>
<p class="p1"><strong>Note: </strong>Accept the argument that Column 2 = Column 1 + Column 3</p>
<p class="p1"> </p>
<p class="p1">(ii) to find the kernel \(\left( {\begin{array}{*{20}{c}} 1&2&1 \\ 0&3&3 \\ 0&0&0 \\ 0&0&0 \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \\ 0 \\ 0 \end{array}} \right)\) <strong><em>M1</em></strong></p>
<p class="p1"><strong>Note: </strong>Allow the use of the original matrix</p>
<p class="p1"> </p>
<p class="p1">\(x + 2y + z = 0\)</p>
<p class="p1">\(3y + 3z = 0\) <strong><em>A1</em></strong></p>
<p class="p1">let \(z = \lambda \) <strong><em>M1</em></strong></p>
<p class="p1">hence \(y = - \lambda ,{\text{ }}x = \lambda \)</p>
<p class="p1">the kernel is therefore \(\left[ {\begin{array}{*{20}{c}} \lambda \\ { - \lambda } \\ \lambda \end{array}} \right]\) <strong><em>A1</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>column rank is \(2\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>a basis for the range is defined by two independent vectors <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1">therefore a basis for the range is for example, \(\left[ {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 3} \\ 1 \end{array}} \right]\) and \(\left[ {\begin{array}{*{20}{c}} 2 \\ 7 \\ 1 \\ 5 \end{array}} \right]\) <span class="Apple-converted-space"> </span><strong><em>A2</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">Many solutions to this question suggested that the topic had not been adequately covered in many centres so that solutions were either good or virtually non existent. Most successful candidates used their calculator to perform the row reduction.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>Many solutions to this question suggested that the topic had not been adequately covered in many centres so that solutions were either good or virtually non existent. Most successful candidates used their calculator to perform the row reduction.</p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Let <em><strong>S</strong></em> be the set of matrices given by</span></p>
<p style="text-align: center;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\left[ \begin{array}{l}<br>a\\<br>c<br>\end{array} \right.\left. \begin{array}{l}<br>b\\<br>d<br>\end{array} \right]\) ; \(a,b,c,d \in \mathbb{R}\), \(ad - bc = 1\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The relation \(R\) is defined on \(S\) as follows. Given \(\boldsymbol{A}\) , </span><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: times new roman,times; font-size: medium;">\(\boldsymbol{B} \in S\)</span> , \(\boldsymbol{ARB}\) if and only if there </span><span style="font-size: medium;"><span style="font-family: times new roman,times;">exists </span></span><span style="font-size: medium;"><span style="font-family: times new roman,times;"><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: times new roman,times; font-size: medium;">\(\boldsymbol{X} \in S\)</span></span> </span></span><span style="font-size: medium;"><span style="font-family: times new roman,times;">such that </span></span><span style="font-family: times new roman,times; font-size: medium;">\(\boldsymbol{A} = \boldsymbol{BX}\)</span><span style="font-size: medium;"><span style="font-family: times new roman,times;"> .<br></span></span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Show that \(R\) is an equivalence relation.</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The relationship between \(a\) , \(b\) , \(c\) and \(d\) is changed to \(ad - bc = n\) . State, with </span><span style="font-family: times new roman,times; font-size: medium;">a reason, whether or not there are any non-zero values of \(n\) , other than \(1\), </span><span style="font-family: times new roman,times; font-size: medium;">for which \(R\) is an equivalence relation.</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-size: medium; font-family: times new roman,times;">since \(\boldsymbol{A} = \boldsymbol{AI}\) where \(\boldsymbol{I}\) is the identity <strong><em>A1 </em></strong></span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">and \(\det (\boldsymbol{I}) = 1\) , <em><strong>A1</strong> </em></span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">\(R\) is reflexive </span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">\(\boldsymbol{ARB} \Rightarrow \boldsymbol{A} = \boldsymbol{BX}\) where \(\det (\boldsymbol{X}) = 1\) <strong><em>M1</em> </strong></span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">it follows that \(\boldsymbol{B} = \boldsymbol{A}{\boldsymbol{X}^{ - 1}}\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">and \(\det ({\boldsymbol{X}^{ - 1}}) = \det{(\boldsymbol{X})^{ - 1}} = 1\) <em><strong>A1</strong> </em></span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">\(R\) is symmetric </span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">\(\boldsymbol{ARB}\) and \(\boldsymbol{BRC} \Rightarrow \boldsymbol{A} = \boldsymbol{BX}\) and \(\boldsymbol{B} = \boldsymbol{CY}\) where \(\det (\boldsymbol{X}) = \det (\boldsymbol{Y}) = 1\) <strong><em>M1</em> </strong></span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">it follows that \(\boldsymbol{A} = \boldsymbol{CYX}\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">\(\det (\boldsymbol{YX}) = \det (\boldsymbol{Y})\det (\boldsymbol{X}) = 1\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">\(R\) is transitive </span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">hence \(R\) is an equivalence relation <strong><em>AG </em></strong></span></p>
<p><span style="font-size: medium; font-family: times new roman,times;"><strong><em>[8 marks]</em> </strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">for reflexivity, we require \(\boldsymbol{ARA}\) so that \(\boldsymbol{A} = \boldsymbol{AI}\) (for all \(\boldsymbol{A} \in S\) ) <strong><em>M1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">since \(\det (\boldsymbol{I}) = 1\) and we require \(\boldsymbol{I} \in S\) the only possibility is \(n = 1\) <strong><em>A1 </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[2 marks] </span></em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question was not well done in general, again illustrating that questions involving both matrices and equivalence relations tend to cause problems for candidates. A common error was to assume, incorrectly, that ARB and BRC \( \Rightarrow A = BX\) and \(B = CX\) , not realizing that a different "\(x\)" is required each time. In proving that \(R\) is an equivalence relation, consideration of the determinant is necessary in this question although many candidates neglected to do this. </span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">In proving that \(R\) is an equivalence relation, consideration of the determinant is necessary in this question although many candidates neglected to do this. </span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">The matrix <strong><em>M </em></strong>is defined by <strong><em>M</em></strong> = \(\left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\).</span><span style="background-color: #f7f7f7; line-height: normal;"><span style="font-family: 'times new roman', times; font-size: medium;"><br></span></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="background-color: #f7f7f7; line-height: normal;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">The eigenvalues of <strong><em>M </em></strong>are denoted by \({\lambda _1},{\text{ }}{\lambda _2}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Show that \({\lambda _1} + {\lambda _2} = a + d\) and \({\lambda _1}{\lambda _2} = \det \)(<strong><em>M</em></strong>).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Given that \(a + b = c + d = 1\), show that 1 is an eigenvalue of <strong><em>M</em></strong>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(c) Find eigenvectors for the matrix \(\left( {\begin{array}{*{20}{c}}2&{ - 1}\\3&{ - 2}\end{array}} \right)\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) the eigenvalues satisfy</span><span style="font-family: 'times new roman', times; font-size: medium;"> \(\left| {\begin{array}{*{20}{c}} {a - \lambda }&b \\ c&{d - \lambda } \end{array}} \right| = 0\) <em><strong>(M1)</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\lambda ^2} - (a + d)\lambda + ad - bc = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">using the sum and product properties of the roots of a quadratic equation <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\lambda _1} + {\lambda _2} = a + d,{\text{ }}{\lambda _1}{\lambda _2} = ad - bc = \det \,\)(<strong><em>M</em></strong>) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) let \(f(\lambda ) = {\lambda ^2} - (a + d)\lambda + ad - bc\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">putting \(b = 1 - a\) and \(d = 1 - c\), consider <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(f(1) = 1 - a - 1 + c + a - ac - c + ac = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">therefore \(\lambda = 1\) is an eigenvalue <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Allow substitution for \(b\), \(c\) into the quadratic equation for \(\lambda \) followed by solution of this equation.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) using any valid method <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">the eigenvalues are 1 and –1 <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">an eigenvector corresponding to \(\lambda = 1\) satisfies</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}2&{ - 1}\\3&{ - 2}\end{array}} \right)\left( \begin{array}{l}x\\y\end{array} \right) = \left( \begin{array}{l}x\\y\end{array} \right)\) <strong>or</strong> \(\left( {\begin{array}{*{20}{c}}1&{ - 1}\\3&{ - 3}\end{array}} \right)\left( \begin{array}{l}x\\y\end{array} \right) = \left( \begin{array}{l}0\\0\end{array} \right)\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( \begin{array}{l}x\\y\end{array} \right) = \left( \begin{array}{l}1\\1\end{array} \right)\) <strong>or </strong>any multiple <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">an eigenvector corresponding to \(\lambda = - 1\) satisfies</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}2&{ - 1}\\3&{ - 2}\end{array}} \right)\left( \begin{array}{l}x\\y\end{array} \right) = - \left( \begin{array}{l}x\\y\end{array} \right)\) <strong>or</strong> \(\left( {\begin{array}{*{20}{c}}3&{ - 1}\\3&{ - 1}\end{array}} \right)\left( \begin{array}{l}x\\y\end{array} \right) = \left( \begin{array}{l}0\\0\end{array} \right)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( \begin{array}{l}x\\y\end{array} \right) = \left( \begin{array}{l}1\\3\end{array} \right)\) <strong>or</strong> any multiple <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Award <strong><em>M1A1A1 </em></strong>for calculating the first eigenvector and <strong><em>M1A1 </em></strong>for the second irrespective of the order in which they are calculated.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[7 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The matrix \(\boldsymbol{A}\) is given by \[\boldsymbol{A} = \left( {\begin{array}{*{20}{c}}<br>0&1&0\\<br>2&4&1\\<br>4&{ - 11}&{ - 2}<br>\end{array}} \right) .\]<br></span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Find the matrices \({\boldsymbol{A}^2}\) and </span><span style="font-family: times new roman,times; font-size: medium;">\({\boldsymbol{A}^3}\)</span><span style="font-family: times new roman,times; font-size: medium;"> , and verify that \({{\boldsymbol{A}}^3} = 2{{\boldsymbol{A}}^2} - {\boldsymbol{A}}\)</span><span style="font-family: times new roman,times; font-size: medium;"> .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Deduce that \({{\boldsymbol{A}}^4} = 3{{\boldsymbol{A}}^2} - 2{\boldsymbol{A}}\)</span><span style="font-family: times new roman,times; font-size: medium;"> .</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Suggest a similar expression for </span><span style="font-family: times new roman,times; font-size: medium;">\({\boldsymbol{A}^n}\)</span><span style="font-family: times new roman,times; font-size: medium;"> in terms of \(\boldsymbol{A}\) and </span><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: times new roman,times; font-size: medium;">\({\boldsymbol{A}^2}\)</span> , valid for \(n \ge 3\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Use mathematical induction to prove the validity of your suggestion.</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) \({{\boldsymbol{A}}^2} = \left( {\begin{array}{*{20}{c}}<br> 2&4&1 \\ <br> 4&7&2 \\ <br> { - 14}&{ - 26}&{ - 7} <br>\end{array}} \right)\)</span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({{\boldsymbol{A}}^3} = \left( {\begin{array}{*{20}{c}}<br> 4&7&2 \\ <br> 6&{10}&3 \\ <br> { - 24}&{ - 41}&{ - 12} <br>\end{array}} \right)\)</span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(2{{\boldsymbol{A}}^2} - {\boldsymbol{A}} = 2\left( {\begin{array}{*{20}{c}}<br> 2&4&1 \\ <br> 4&7&2 \\ <br> { - 14}&{ - 26}&{ - 7} <br>\end{array}} \right) - \left( {\begin{array}{*{20}{c}}<br> 0&1&0 \\ <br> 2&4&1 \\ <br> { - 4}&{ - 11}&{ - 2} <br>\end{array}} \right)\) <strong><em>M1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = \left( {\begin{array}{*{20}{c}}<br> 4&7&2 \\ <br> 6&{10}&3 \\ <br> { - 24}&{ - 41}&{ - 12} <br>\end{array}} \right) = {{\boldsymbol{A}}^3}\)</span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>AG</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \({{\boldsymbol{A}}^4} = {\boldsymbol{A}}{{\boldsymbol{A}}^3}\)</span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>M1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = {\boldsymbol{A}}(2{{\boldsymbol{A}}^2} - {\boldsymbol{A}})\)</span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> \( = 2{{\boldsymbol{A}}^3} - {{\boldsymbol{A}}^2}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = 2(2{{\boldsymbol{A}}^2} - {\boldsymbol{A}}) - {{\boldsymbol{A}}^2}\)</span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = 3{{\boldsymbol{A}}^2} - 2{\boldsymbol{A}}\)</span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>AG</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Accept alternative solutions that include correct calculation of both </span><span style="font-family: times new roman,times; font-size: medium;">sides of the expression.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[6 marks]</span></strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) conjecture: \({{\boldsymbol{A}}^n} = \left( {n - 1} \right){{\boldsymbol{A}}^2} - \left( {n - 2} \right){\boldsymbol{A}}\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) first check that the result is true for \(n = 3\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the formula gives \({{\boldsymbol{A}}^3} = 2{{\boldsymbol{A}}^2} - {\boldsymbol{A}}\) which is correct <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">assume the result for \(n = k\) , <em>i.e.</em> <strong><em> M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({{\boldsymbol{A}}^k} = (k - 1){{\boldsymbol{{\rm A}}}^2} - (k - 2){\boldsymbol{A}}\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">so</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({{\boldsymbol{A}}^{k + 1}} = {\boldsymbol{A}}\left[ {\left( {k - 1} \right){{\boldsymbol{A}}^2} - \left( {k - 2} \right){\boldsymbol{A}}} \right]\) <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> \( = \left( {k - 1} \right){{\boldsymbol{A}}^3} - \left( {k - 2} \right){{\boldsymbol{A}}^2}\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> \( = \left( {k - 1} \right)\left( {2{{\boldsymbol{A}}^2} - {\boldsymbol{A}}} \right) - \left( {k - 2} \right){{\boldsymbol{A}}^2}\) <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> \( = k{{\boldsymbol{A}}^2} - \left( {k - 1} \right){\boldsymbol{A}}\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">so true for \(n = k \Rightarrow \) true for \(n = k + 1\) and since true for \(n = 3\) ,</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the result is proved by induction <em><strong>R1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Only award the <em><strong>R1</strong></em> mark if a reasonable attempt at a proof by </span><span style="font-family: times new roman,times; font-size: medium;">induction has been made.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[8 marks]</span></strong></em></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">Consider the system of equations \[\left( {\begin{array}{*{20}{c}}<br>1&{ - 1}&2\\<br>2&2&{ - 1}\\<br>3&5&{ - 4}\\<br>3&1&1<br>\end{array}} \right)\left( \begin{array}{l}<br>x\\<br>y\\<br>z<br>\end{array} \right) = \left( \begin{array}{l}<br>5\\<br>3\\<br>1\\<br>k<br>\end{array} \right) .\]<br></span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">By reducing the augmented matrix to row echelon form,</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) find the rank of the coefficient matrix;</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (ii) find the value of \(k\) for which the system has a solution.</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">For this value of \(k\) , determine the solution.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">reducing to row echelon form</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\begin{array}{*{20}{ccc|c}}<br> 1&{ - 1}&2&5 \\ <br> 0&4&{ - 5}&{ - 7} \\ <br> 0&8&{ - 10}&{ - 14} \\ <br> 0&4&{ - 5}&{k - 15} <br>\end{array}\) <em><strong>(M1)(A1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\begin{array}{*{20}{ccc|c}}<br> 1&{ - 2}&2&5 \\ <br> 0&4&{ - 5}&{ - 7} \\ <br> 0&0&0&0 \\ <br> 0&0&0&{k - 8} <br>\end{array}\) <em><strong>A1</strong></em><br></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) this shows that the rank of the matrix is \(2\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) the equations can be solved if \(k = 8\) <strong><em>A1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[5 marks]</span></em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">let \(z = \lambda \) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">then \(y = \frac{{5\lambda - 7}}{4}\) </span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">and \(x = \left( {5 - 2\lambda + \frac{{5\lambda - 7}}{4} = } \right)\frac{{13 - 3\lambda }}{4}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Accept equivalent expressions.</span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></strong></em></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Show that the following vectors form a basis for the vector space \({\mathbb{R}^3}\) .\[\left( \begin{array}{l}<br>1\\<br>2\\<br>3<br>\end{array} \right);\left( \begin{array}{l}<br>2\\<br>3\\<br>1<br>\end{array} \right);\left( \begin{array}{l}<br>5\\<br>2\\<br>5<br>\end{array} \right)\]</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Express the following vector as a linear combination of the above vectors.\[\left( \begin{array}{l}<br>12\\<br>14\\<br>16<br>\end{array} \right)\]</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">let</span><span style="font-family: times new roman,times; font-size: medium;"> \({\boldsymbol{A}} = \left| {\begin{array}{*{20}{c}}<br> 1&2&5 \\ <br> 2&3&2 \\ <br> 3&1&5 <br>\end{array}} \right|\) and consider \(\det (\boldsymbol{A}) = - 30\) <em><strong>(M1)A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the vectors form a basis because the determinant is non-zero (or because </span><span style="font-family: times new roman,times; font-size: medium;">the matrix is non-singular) <strong><em>R1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">let \(\left( \begin{array}{l}<br>12\\<br>14\\<br>16<br>\end{array} \right) = \lambda \left( \begin{array}{l}<br>1\\<br>2\\<br>3<br>\end{array} \right) + \mu \left( \begin{array}{l}<br>2\\<br>3\\<br>1<br>\end{array} \right) + v\left( \begin{array}{l}<br>5\\<br>2\\<br>5<br>\end{array} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1A1</span></strong></em></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">so that</span></p>
<p align="LEFT"><strong><span style="font-family: times new roman,times; font-size: medium;">EITHER</span></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\begin{array}{l}<br>\lambda + 2\mu + 5v = 12\\<br>2\lambda + 3\mu + 2v = 14\\<br>3\lambda + \mu + 5v = 16<br>\end{array}\) <em><strong></strong></em></span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1</span></strong></em></p>
<p align="LEFT"><strong><span style="font-family: times new roman,times; font-size: medium;">OR</span></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}<br>1&2&5\\<br>2&3&2\\<br>3&1&5<br>\end{array}} \right)\left( \begin{array}{l}<br>\lambda \\<br>\mu \\<br>v<br>\end{array} \right) = \left( \begin{array}{l}<br>12\\<br>14\\<br>16<br>\end{array} \right)\) <em><strong>M1</strong></em></span></p>
<p align="LEFT"><strong><span style="font-family: times new roman,times; font-size: medium;">THEN</span></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">giving \(\lambda = 3\), \(\mu = 2\), \(v = 1\) <em><strong>(A1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">hence \(\left( \begin{array}{l}<br>12\\<br>14\\<br>16<br>\end{array} \right) = 3\left( \begin{array}{l}<br>1\\<br>2\\<br>3<br>\end{array} \right) + 2\left( \begin{array}{l}<br>2\\<br>3\\<br>1<br>\end{array} \right) + 1\left( \begin{array}{l}<br>5\\<br>2\\<br>5<br>\end{array} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1</span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[5 marks]</span></strong></em></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The set \(S\) contains the eight matrices of the form\[\left( {\begin{array}{*{20}{c}}<br>a&0&0\\<br>0&b&0\\<br>0&0&c<br>\end{array}} \right)\]where \(a\), \(b\), \(c\) can each take one of the values \( + 1\) or \( - 1\) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Show that any matrix of this form is its own inverse.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Show that \(S\) forms an Abelian group under matrix multiplication.</span></p>
<div class="marks">[9]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Giving a reason, state whether or not this group is cyclic.</span></p>
<div class="marks">[1]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}<br>a&0&0\\<br>0&b&0\\<br>0&0&c<br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br>a&0&0\\<br>0&b&0\\<br>0&0&c<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>{{a^2}}&0&0\\<br>0&{{b^2}}&0\\<br>0&0&{{c^2}}<br>\end{array}} \right)\)</span><span style="font-family: times new roman,times; font-size: medium;"> <em><strong>A1M1</strong></em></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( = \left( {\begin{array}{*{20}{c}}<br>1&0&0\\<br>0&1&0\\<br>0&0&1<br>\end{array}} \right)\)</span><span style="font-family: times new roman,times; font-size: medium;"> <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">this shows that each matrix is self-inverse</span></p>
<p><em><strong> <span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">closure:</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}<br>{{a_1}}&0&0\\<br>0&{{b_1}}&0\\<br>0&0&{{c_1}}<br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br>{{a_2}}&0&0\\<br>0&{{b_2}}&0\\<br>0&0&{{c_2}}<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>{{a_1}{a_2}}&0&0\\<br>0&{{b_1}{b_2}}&0\\<br>0&0&{{c_1}{c_2}}<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1A1</span></strong></em></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> \( = \left( {\begin{array}{*{20}{c}}<br>{{a_3}}&0&0\\<br>0&{{b_3}}&0\\<br>0&0&{{c_3}}<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">where each of \({a_3}\), \({b_3}\), \({c_3}\) can only be \( \pm 1\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">this proves closure</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">identity: the identity matrix is the group identity <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">inverse: as shown above, every element is self-inverse <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">associativity: this follows because matrix multiplication is associative <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(S\) is therefore a group <em><strong>AG</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Abelian:</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}<br>{{a_2}}&0&0\\<br>0&{{b_2}}&0\\<br>0&0&{{c_2}}<br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br>{{a_1}}&0&0\\<br>0&{{b_1}}&0\\<br>0&0&{{c_1}}<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>{{a_2}{a_1}}&0&0\\<br>0&{{b_2}{b_1}}&0\\<br>0&0&{{c_2}{c_1}}<br>\end{array}} \right)\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}<br>{{a_1}}&0&0\\<br>0&{{b_1}}&0\\<br>0&0&{{c_1}}<br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br>{{a_2}}&0&0\\<br>0&{{b_2}}&0\\<br>0&0&{{c_2}}<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>{{a_1}{a_2}}&0&0\\<br>0&{{b_1}{b_2}}&0\\<br>0&0&{{c_1}{c_2}}<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note:</strong> Second line may have been shown whilst proving closure, however a r</span><span style="font-family: times new roman,times; font-size: medium;">eference to it must be made here.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">we see that the same result is obtained either way which proves commutativity </span><span style="font-family: times new roman,times; font-size: medium;">so that the group is Abelian <strong><em>R1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> [9 marks]</span></em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">since all elements (except the identity) are of order \(2\), the group is not cyclic </span><span style="font-family: times new roman,times; font-size: medium;">(since \(S\) contains \(8\) elements) <strong><em>R1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[1 mark]</span></em></strong></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p>By considering the images of the points (1, 0) and (0, 1),</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>determine the 2 × 2 matrix <em><strong>P</strong></em> which represents a reflection in the line \(y = \left( {{\text{tan}}\,\theta } \right)x\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>determine the 2 × 2 matrix <em><strong>Q</strong></em> which represents an anticlockwise rotation of <em>θ</em> about the origin.</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Describe the transformation represented by the matrix <em><strong>PQ</strong></em>.</p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>A matrix <em>M</em> is said to be orthogonal if <strong><em>M </em></strong><em><sup>T</sup></em><strong><em>M</em></strong> = <strong><em>I</em></strong> where <strong><em>I</em></strong> is the identity. Show that <em><strong>Q</strong></em> is orthogonal.</p>
<div class="marks">[2]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><img src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAcwAAAF1CAYAAABlHto7AAAgAElEQVR4Ae2dDYhV13q/l/cfW3OJFFOci5FIMWBgEmJAQcsoWlAwoFJbFCIYiRcM6MVILZ2ACSokoKAXlUaYwDWo1ICBGoxUwdAqKokwgZt0lKvgLVUmUgUDUSblKj1/npWs4549+5zZ+5z9ufZvwczZZ3+sj2fts9/9vutd75rQaDQaRkkEREAEREAERKAtgV+0PaqDIiACIiACIiACloAEpm4EERABERABEYhBQAIzBiSdIgIiIAIiIAISmLoHREAEREAERCAGAQnMGJB0igiIgAiIgAhIYOoeEAEREAEREIEYBCQwY0DSKSIgAiIgAiIggal7QAREQAREQARiEJDAjAFJp4iACIiACIiABKbuAREQAREQARGIQUACMwYknVIfAt9//7354x//aD799NNRjd6zZ4/dP2qnvoiACNSKgARmrbq7fo1FACZJCMsvvvjCrFmzxrhrv/76a/POO++Yzz//PElWOlcERMAzAhKYnnWomvOEwNWrV83atWvNjz/++GTnOFtz5swxGzduNHwiPElsDwwMmJkzZ45ztQ6LgAj4TEAC0+ferXnbtm7das6ePWtOnz6dmMS0adOaApOL0TIXLFiQOB9dIAIi4A8BCUx/+lItCRDAhIqQQyt86623mubVwCltN5977rmmwCQvtM4pU6a0vUYHRUAE/CYggel3/9aydd988405evSoOXLkiDWn9vX1WaH56NGj2DyeeuopKzC/+uorw/WYZZVEQATqTUACs97972XrDx8+bHbu3GlWrFhhJk2aZA4cOGC9Xm/cuBG7vVOnTrUaKteQj5IIiIAISGDqHvCKAI4+t27dsiZUGva///u/1ix76tQpc+jQodhtnTx5sj131apVsa/RiSIgAn4TeMrv5ql1dSKAV+vLL79sBgcHxzQbLfGjjz4as7/VDgRmVD6tztd+ERAB/wlIw/S/j2vTQoIL4JjTarwRRx7mWI6XELyaQjIeJR0XgfoRkIZZvz73ssUIQgILDA0NjWofzjsubdq0ycyfP9/893//t+np6XG77SdzNTmGCZbpI0uWLBl1XF9EQAREYEKj0WgIgwhUncDcuXMjTahE7Dlx4kSzeUwRQQvt7+9v7nMbRPbR1BFHQ58iIAJhAjLJhonoe+UIIOjwho2TVq9ebXAMikoSllFUtE8ERMARkMB0JPRZWQKYY3fv3h2r/oxvytwaC5VOEgERCBGQSTYERF/9IhA2yfrVOrVGBEQgTwLSMPOkrbJEQAREQAQqS0ACs7Jdp4qLgAiIgAjkSUACM0/aKksEREAERKCyBCQwK9t1qrgIiIAIiECeBCQw86StskRABERABCpLQAKzsl2niouACIiACORJQAIzT9oqK3cCwdB4uReuAkVABLwiIIHpVXeqMWECEphhIvouAiLQKQEJzE7J6bpKEGA9TCUREAERSIOABGYaFJVHaQlIwyxt16hiIlA5AhKYlesyVTgJgcePHyc5XeeKgAiIQEsCEpgt0eiACIiACIiACDwhIIH5hIW2REAEREAERKAlAQnMlmh0wAcCcdfJ9KGtaoMIiEC2BCQws+Wr3AsmIC/ZgjtAxYuARwQkMD3qTDVlLAFpmGOZaI8IiEBnBCQwO+OmqypCQBpmRTpK1RSBChCQwKxAJ6mKIiACIiACxROQwCy+D1QDERABERCBChCQwKxAJ6mKnRPQGGbn7HSlCIjAaAISmKN56JtnBDSG6VmHqjkiUCABCcwC4avo7AlIw8yesUoQgboQkMCsS0/XtJ3SMGva8Wq2CGRAQAIzA6jKUgREQAREwD8CEpj+9alaJAIiIAIikAEBCcwMoCrL8hDQepjl6QvVRASqTkACs+o9qPq3JaD1MNvi0UEREIEEBCQwE8DSqdUjIC/Z6vWZaiwCZSUggVnWnlG9UiEgL9lUMCoTERABY4wEpm4DERABERABEYhBQAIzBiSdIgIiIAIiIAISmLoHvCYgL1mvu1eNE4FcCUhg5opbheVNQF6yeRNXeSLgLwEJTH/7Vi0zxshLVreBCIhAWgQkMNMiqXxKSSCOl+yPP/5oPv/8c/Pdd98128D2p59+aq5evdrc1+0G5Vy/fn1UNpcuXRqzb9QJ+iICIlAaAhKYpekKVaQIAgjEhQsXmgULFpjnnnvOVuGLL74wa9asMUuWLDEHDx4077zzTldVQ/iS3y9/+Uvz3nvvjcqLcj/++GMrnEcd0BcREIHSEXiqdDVShUQgRQLtnH7Q+Pbt22cGBgbMlClTbKlofG+99ZbZvn273bdx40YrTDne39/fUc2oA9cGNdhgRpTxxhtv2ONvv/128JC2RUAESkRAGmaJOkNVSZ9AO6ef06dPm4cPH5o5c+Y0C961a5f54x//aBYvXmz39fb2mqefftpqmQjYTlJPT48t46WXXoocU505c6YVmFu3brVld1KGrhEBEciegARm9oxVQkkJoEkiEF36/vvvzddff22/Oo0TYekE6rVr19ypHX+2GlNdvXq1zZOxVCUREIFyEpDALGe/qFYpEWhlksWhBwE5a9asZkl85w/P2uB1ztOWY1klBHRfX5+5cuVKVkUoXxEQgS4JSGB2CVCXl5tAK5PsRx99ZCuOudQlZ3JFqwwKTL6T3HF3ftqfmGxv3bqVdrbKTwREICUCEpgpgVQ21STgTK/U3gnGcEucZhk8N3xO3O9OW406n/xdWVHHtU8ERKBYAhKYxfJX6RkTePDgQWQJTlt89OhR8/jUqVOtZyxCy13Hcefd6qadNC/oYKOVxuuyCmq2bp8+RUAEykFAArMc/aBaZEQAEyder+HkhB9esi5NnjzZOOebe/fu2d0IOJx90P7wZu02tROYCGrqoCQCIlBOAhKY5ewX1SpFAnv27BmTmxOMTnt0J+zevdsKx8uXL9tdzmv2xIkT7hT7ScADghEQ5CCthGBPQyinVR/lIwIiMJqABOZoHvrmGQE0NqZqhLVMovjglRrejyaJcCSE3dGjR61AJLAB5wcTAhVh6QIOBI+Ft9EcXeg9TLxc50zC7lwEN8KZyD9KIiAC5SQwodFoNMpZNdVKBLongEBDuKFRoj0GE4LrwIEDVpgF97ONkEOAMU/TmW+D53AcYctcTgSqm6sZPMdtIxzv3LljNVeXd1iTxGuXvIaHhyPLc3npUwREoDgCEpjFsVfJORBAYBJ6jnixP/zww5gxQuLEommuWLEiUW0Qggg5BGFYECfKyBgreCkfzZapJUoiIALlJCCTbDn7RbVKkQBmTkyqJ0+eHJMrwg6nHrS7sJl0zMmBHWiXCONuhSWaKqZfxkQlLAOAtSkCJSQggVnCTlGV0iPgQtEhMBFOUYnA6JhI0RbjJoRbGvMyMfuGx0fj1kHniYAI5EtAJtl8eau0nAngyeo8XDF7KlZrzh2g4kTAIwLSMD3qTDWlPQFF0mnPR0dFQATaE5DAbM9HRz0igMBMc96kR2jUFBEQgRgEJDBjQNIpfhDAJItXbDAcnh8tUytEQATyICCBmQdllVEKAs7x59tvvy1FfVQJERCBahGQwKxWf6m2XRIgwABTOJREQAREICkBCcykxHR+pQns37/fnD9/vtJtUOVFQASKISCBWQx3lVoQAeZP3r59u6DSVawIiECVCTxV5cqr7iLQCYFXXnnFzsdMGg6vk7J0zWgCZ86cMYsXL265WPfos9P7RpSnb775xmY4a9Yss3bt2sjMCW7BuTNmzLCme+IIu5VtIi/ocOe2bdvMhg0buoruxNDCp59+amvAIgOvv/56ZBxiIljRpp6eHhuGcdq0aTbIPx7jDFGE4xp32KR6XEbwdSUR8JXA6tWrxzTt1KlTjTlz5ozZrx3ZEujv728cOXIk20Iich8cHGxMmjSJRSaaf+fOnYs4s9Hgftm9e7c9dvPmzcbMmTMjz+t2Z29vb6NVHeLm3dfX12wPbVuyZEnkpbTJcb9//37z3qd9W7ZsibxGO6MJyCRbj/citTJAgLdqQtLxp5QPASIuwZv4u3knyr506ZJhYab79+9brWrp0qVjlnZj3VTqSKhEktO8wkvApVF/tMNuQiKiWWIhoU38bdmyxc4xpq0uoVk+++yzNoSj485cZBczmfbxPWq9WJeHPkcTkMAczUPfakBg6tSp9mGlB0U+nY2wIiRhq0D1rAXqHuLhGmEibXWcazjGOa3m1j548MAKQLf8GgLizTfftMUEX5gQigghF0aRE8if67tJrn7h9oW/U4aLdcyxdm1y9XHt4Dsr8pAQxI7F4cOHrdDfsWOHu8S26d69e82yMOPu3LnTLgDQPEkbrQlEK57aKwJ+EIgyydIyzHSYsTBRKWVLALNflLlwZGSkgZkW0yLHT5w4MaoiFy9etPsxn0f1I9dyHcd27Ngx6tp2XzDJ0/dBk+jGjRvH1NGZcju9RwYGBqz5c/ny5Q3yd+VhCuU7nyTyd21hH+2dMmVKg3rGTeRBm4JDDZh9w0zJn7yDCbMz++gPpfYEpGG2fpfQEY8JoHHwh/ajlC0Bt4B3uJT169eP6gPMiS50IdofUZnOnTtnBgcHDY4qEyZMsBol+WAdYB/H0QqD2mK4nPB3tEnMoc4kyrWsbcq6qMGExunMlsH9cba5lrZQd+rHNlojfxyjPJf47szBaHto4tRt69at7pRxP9FkSc6cTJtYqSfssES5ztTsMqUs6sUyd0rtCUhgtuejox4T4OGiOZnZdjDmRR7mCLdg4oHOQ9o90OmLdevWmUmTJtnzWZ806MW8adMmewwzI4kXHbxeyYPE8biJsoPmYYQU6emnn7amY8zHBw4csEJt9uzZcbMddd6tW7fsGCl5PfXUU2ZgYMCOF2ISpq3PP/9883zXblht377dCkvmC9+9e7elqbp58c8bZ8+eNZhely9fbvfQJpjDCWHNH+uu8ufM0y4PJ0CvX7/udumzFYH2CqiOikC1CUSZ8oItwlymlB0BTIB4qDpzpCsJE2SrvsGUiXkxbE7E3Oj6y53DvnDeroyoT/J0XrAcd6ZMyguaJF3+fHaSnNmXfPfv3z8mC8ygziTLQUyiYbMo3+Oag4Msh4eHLb9wua5NYa5uf/j8MZXWjoY0zFZvEtpfCwJoFU5LqUWDS9TIKMcXquf6I3wcTcgtCI6TC1oY2iIer84LtF3zMHtiknRmS851HrCYJbkXSJSLOZjkNDb7JcE/tGPMxWhzmFbxVqWuWaS5c+eOapNbCJ0gHS4F2+RM0e4YWq9SPAISmPE46SxPCSxatMh6CXravMKb5cYAo4Qf5nA39kZFOQcTphtLdIEGXCM4l6ATJEyLmGGZUoFgYmI+ZsdWyY3dca5LCEvnBRsUjM7sOzQ0FBkIwF3f7hPhiGBiDJM/0oIFC5ovA62uxXwbN8GHFwXGSINmVsc0aPal/QjGmzdv2s9gGe6lgSANSu0JSGC256OjnhN47bXXzMGDB5uahufNLaR5PMxx/AkmNESi6UyfPt06vDBmyTZaEYKFMb+PP/64KWCYLkEeRMchHTt2zHz22Wd2G8E0f/78Mc4srjy0RcYkca7BsYg/tDI+ifpDcloZwuPQoUN2vmZQQ+McBDKORy+88EKzXq6M8Cdjh05Lpf1Og3XaHC8HTpPmWrb5e/z4cTOr8EtG88DPmjH3Lho35dAWNG3q5zTlhw8f2kto0+nTp62wdOOVwbyc5tvb2xvcre0IAv9vpxvxjjioXSJQdQJoHn//93/fshnPPPOM+fd//3eDkwYPIKX0CTDvD+GHkJw4cWKzAB7ev/jFL6x35q9+9StDuDgEH8lpTAge+ua//uu/rDZFWD0SwsXNJ2S5NvIKmxo5D2GBoEM4I0gQWHz+1V/9lRW+CE723bhxwworhIcT5ragwD80N8pEqOEMRB6tEoLsP//zP+351OH//u//bPvQ+r788kubx5/92Z8Z5gST75/+9CerWfP9L//yL82VK1csK66jfoS+CyY0bELd/fmf/3mzTX/xF39h8+ClwuX7ww8/WFMwmmiUBomG/U//9E82XGG4f4LlafsnAhMYxxUMEfCVAG/ewcnoUe10msfIyEjz7TzqPO3rjAACAe2RfnBesZ3lVI6r0OgQzlECuhw1jF8LXhB4aSACktN+419dvzNlkq1fn6vFIQI8xNFQwmbD0Gn62iEBNBvGDjG7uvG1DrMq/DLGAtFufRCWwERTZYxVwjLerSWBGY+TzqoogbhOFMTixPynlA0BBMyRI0esebTd2Fw2paeTK8IewRKcw5lOzvnnQlswVWPidubv/GtRvRJlkq1en6nGCQjEMcmSHVoD4zx4aSplRwDOJGk02TGOkzPjqqQoJ6A419f1nPg+zHUlpHbXgoAe4Pl0szjnw3m8UiQoxyMUfVwm2Wgu2usJgbgmWZqL5yNjOkoiIAIiEEVAAjOKivZ5QyA4r228RjGes2vXrvFO03EREIGaEpDArGnHq9ljCeD8wBhbu4gxY6/SnqwIMM7mxjyzKkP5ikASAhKYSWjp3MoRSGKSZXyNEGlEhXGL8FauwR5UGCHJ9I21a9fKc9mD/vSpCRKYPvWm2jKGQBKTLBe//fbbNowYEV2UiiHAfE3+MJETtlBJBMpCQAKzLD2hemRCIImGSQUwy7I+ISH1lPIlgAmWqDNo+QQgY1URzOOaH5tvP6i01gQkMFuz0REPCCTVMGkyJtkLFy540PpqNIFABghKgpoT3MAt1YWJnPimLI+lJAJlICCBWYZeUB1KRYAHtZxN8usStEhimhJBJ7xCCBo/mmdVowPlR1El5UFAAjMPyiqjcgRYVYIlpZSyI8BLCZGYWH/SmWDDpaFhksJrY4bP03cRyIOABGYelFVGYQQmTZrUUdmrVq0y69ev7+haXTQ+AYQlZlgC3483XszamGfPnh0/U50hAhkTkMDMGLCyL5YA6xJ2klhhQ6bATsjFu2bPnj1m2rRpsZb7wizrFniOl7vOEoFsCEhgZsNVuZaEQFIvWVdtHtKMZWrJL0cknc/jx4+bZ5991rL9j//4j1iZEveURaSVRKBoAgq+XnQPqPxMCXTiJesqhCkQhxRf1j507Sri041XYiJnseKkadasWeqLpNB0fuoEpGGmjlQZ+kLAeWj60p4i24EJlpcPAkN0kjCR40mrJAJFEpDALJK+yi41AUyyhMjTlIbOu4mFionYg2Y5MjLSsbaOlo/QVcjCzvtCV3ZPQAKze4bKocQEOvWSdU2aP3++wrM5GAk/mTIyffp0s2/fPrNz507z9NNPJ8zhyeluTFkBJZ4w0Vb+BCQw82euEnMk0KmXrKviokWLbCBwNCWleARghekVL9hz587ZcIPxrmx/FnMyNR+zPSMdzZaABGa2fJV7wQS61TBZVBqHldOnTxfckmoUzzjjwoULbWUJMZimw5QEZjXugXAtfRrSkMAM966+e0WgWw0TMyIP6nfeeccrLlk05vPPP7fBCOCFsEw7KWRh2kSzy4+XTJZoIzgFZnnmNPuQNK3Eh15UGzIlgADgx0+ovHCs00wLrkjmDx48MIcOHTLXr183p06dMitWrMis5jNmzFA/ZEa384zRIvGCvnbtmrXI4Jw1efJkG5iC8Wfm0vqQJDB96EW1IVMC/NgJ4UYQAwnMsahx6mHdyqGhIcP0jywTJvJjx47ZQO1ZlqO84xFAk/z000/tCyVaJCZ4XjD7+vq6cvKKV3r+Z8kkmz9zlZgjgW7HMF1VWUlD8wAdjSefLPTMiwTBCLIWlpTa29trH9BPaqCtIgigTWJupc8RmPw+hoeHzYkTJ6zQ7MYjuoj2xC1TGmZcUjqvkgS6HcN0jWb8TJ6yjoYxZ86csWZYtO9//ud/fnIg4y00TMx/R48eba6bmXGRtc8eLRLtkYW8XUxfQk5iddm7d6+ZN2+el9pkVMdLYEZR0T5vCKSlYQLERZthTKbOiQfor3/9ayu00vSCjcOUcTHGSDEBu4Wm41ynczojgCMXC3jT57wcoUm6ObGd5Vjtq2SSrXb/qfbjEEhLw6QYHhQ4/9Q54S1M8HQ0vLyFpePOGBnmcWn8jki6nwjJF154wfbzrl27DDGVb968aQYHB22fY22pa5rQYOVWJRHwlADRZhhXSSPhBfrXf/3X5uLFi7Vz/kE44dzz8OFD+9DEHFdUwiS7dOlSs23bNsO6pUqdE4Alnq2YW9Ei+WP8EaGIRkngjp6ens4L8OxKmWQ961A1JzsCL774on2IYA7krbsuiYfoypUrrcMNmmXRiQc65lii/khgdt4bjEtiMcCBh7Rjxw4boanOGuR4NGWSHY+QjleaQJpjmIDAHIhXIG/mdUg8TDHBYn4tg7B0zOkHNCOlZAQQklhd6FO8XOlXzK14ORPOUMKyPU9pmO356GjFCaQ5hgkKTJGsmnHlyhW7CkfF8bSsPi8EjNcyVnjkyJFSOtjgqUk9fZ3C0LJzYh6ADf3HH1YC91tgLH7dunV2TD6PqUAxq1uJ0yQwK9FNqmSnBNLWMHkDR7vxXWASHADvSLQPxrLKmKZOnWrngBblfFRGJq5OYXMr3q3ct0rdEZDA7I6fri45AfdWnWY1t2zZYtauXZtmlqXJi/B/CEqEZNn9ARcsWGCXDZPANDZcIEMFBJG4e/eutX709/en5vBWmhu04IpoDLPgDlDx1SOACZD5gLzF+5QImM5KIzxoq+DUtHz5cisgnNOKT30Rpy2MKbvpH0RcYn4qy6nhDEVf1n2+cByGSc+RwExKTOeLQCCIgS8wGONizBJBWRWNzU1/YIyuTsmZW90KOoyrM3VKAjL7u0DzMLNnrBIKJIDp9Pjx46nXgHlreBsyxldlpxPMd0TtQeignVQtuDxaFtrVDz/8YCZOnJh6P5chQ+bAsh4rJle20SQJbr548WJr6ShDHetSB2mYdenpmrbz8ePHmbSc8TMEJZFQqprQKl999VU7nYAHcdWEJdyZj4mn54ULF6raDZH15gUGwcj0D8zkOK/xQsMYMw48HGNYQClfAnL6yZe3SsuZQNpessHqYwKrsjkQzYw2VMUEG2Qf3KYNCHwfEu0gMAbmce5dPFsxucrcWo7elYZZjn5QLTIikIWXrKvq/v37bZQUQuZVKX344YdWK8NphrihVU+uH6rYDgQk/YHFAm3y3XffbQYT4JgLdl7FtvlYZ2mYPvaq2pQLAUyBeJSePXvWEDav7AkTLDFYmTLii0YGc/qBtlUlYZUgqg6RipjXi1BkTFyp/ASkYZa/j1TDEhPAnOnWCCxxNW3VMMHysGYeqW+pKo5XmFpxFsPTlZctxiWLDGTv232QdXskMLMmrPwLJUD4tCwTApN5b2VOaJPEDcWzkmAEmP98SAidnTt3WgHEgtaELCxTgjsmbwQkWjCOOtyPLJPFMQRmWaMolYljmeoigVmm3lBdUieQlZdssKKzZs1qrvgQ3F+GbaZdTJ8+3caDxQxY9RScrI+HLMKfuaNfffWVFZhFmmaJ3UoQBTR5592K2ZX6OeG5YcMGBTiv8E2Y7et3hcGo6n4QyFrDhBIPbSKrlMnblOkHzNvDXFzV9TsRMggg2vDgwQPDyw+eo5iU0cyC8xDZz3ggAjWvFwMEJHWjjtTVOZhRNxcAQqt/+PEcabaCBaSVRMBXAqtXr868acPDwyzC3hgcHMy8rDgFUJ8pU6Y0Zs6c2bh//36cS0p3zokTJxq9vb22HXPmzGmcO3du3Dr29/fbNo97YgonDA0NNTZu3GjrN23atMbu3bsryzoFHLXJQibZ5quDNkSgMwKMTzFPrixjaHjC4nlJFKKqajhobTjDsE4jY35xtHfOZ1wTbS+LhNbu1pJkQW00yeB4ZFVZZ8HK1zxlkvW1Z9UuSyAPkywF4cDB2BWh5np6enKnT7mHDh2yXrCYJKu+lBM833///UQce3t7jQvI3q3naStzqwv0gACXw06i7vHiZAlML7pRjWhFIA+nH8rmAU0w7G+//TaWNtSqvp3u/+1vf2vHLH0Jwo0wunfvXiIcTC3hZQHP1G4EJpokIQ/RctEaEd7kJw0yUXd4ebJMsl52qxpVBAHMoDxk806YCZlfiQnWpxBqxLbFcSlJQvPDLJs04V2LSR3vVsyt5ANP/tDWJSyTEvXzfGmYfvarWvUzgbxMshTHQzbpA76bjjp58qT55JNPrGmw6ibYKA6LFi0y7733nuWaRGChaSIAW13z6NEjc/v2bRtd59atW01NdurUqXbsd9myZWbGjBlRVdK+mhOQwKz5DeB78/MyycKRBzQP6nYP67R4o0Vt3rzZmh990iqDfJg2Qjt5CUnyQoA51zkNBfNjm75hChABzkmYWrEMtBKu4ev1vd4EZJKtd/+r9SkTIJpO1lomY3QvvPCC94sGoykiKJN6H+P8Qwg6lxCSjC9jboUbcycxteKBy3xJCUtHSp/jEZCGOR4hHa80gTxNsoAiiAEP5zfffDP1BY2dtsXqKKdOnfImxF27GwyBSTACxmjjatJomJyPoEVYkp555hmrSZLHK6+8knrftGuDjvlDYAIzTv1pjloiAqMJ4BCD52heiQc0mgwaTJrTDsiXeLB8MvcvzbzzYtNpOU7DxFu1XYINghJPWbcSCMHN4wradnnrmAhAQAJT94HXBPIWmMDEHIgzSdJ5hFEd4ULcMcVi1apVhUxZiapXnvuYE0k/urU7+Y62zTglQpK/yZMn279p06ZZRhzH23VoaMgGPs+zvirLXwIyyfrbt2pZQQQwI7IyRbeJiDULFy60Y2xolXUda2Ms05lZEZZbt261AhPhyGoleCeH2RB9iX1onGwriUAaBCQw06CoPEpLgKDcRSQe8t0ktCYX4i6Jh2g3ZZb1WljgqINJGocevFwxs4aFZLj+mHARmGm8vITz1vd6EpDArGe/16bVbgWJvBuMYwmm2aTCDuGAQGAMzocQd0m5M0eSaEkIOsL9PXz40KBVwnkiAVEAACAASURBVJOwd2iTcTVGpozAUEkE0iIggZkWSeVTSgJFaZizZ8+2psOkAhMT47Fjx0wdnVUYd8TBx03LYRkvtMTxNMlWNx7X8QKiJAJpEdA8zLRIKp9SEihKw0QTco4pccEQvJ1rmB9YF89O2usWXMbkylilmyPJy0OnwtIxRxtFW1USgTQIyEs2DYrKo7QEivCSdTDeeOMNKwAvXbrkdkV+uhB3OLFs2rTJvPjii5Hn+bDTvUSg+blFoZkjSSg65rA+//zzqc6RPHz4sDl//rydy+kDP7WhWAISmMXyV+kZEyhSYKLZoDUNDw+3HHdzTilMmfBZq0RQYmplfBatHw18x44dhgDrWSam5bz88stmZGTEdOuIlWU9lXc1CMgkW41+Ui07JFDUGCbVRQAiGFqtYIKQRKD6siRXuIsQkoxJEo6OdvJywNgsJmfanLWwpD6UgVlXZtlw7+h7JwTk9NMJNV1TGQJFjWE6QDitIBgxz7rE/Eoce3wLcUd7zp49azVITK54tyKsYMCLAybnIrQ8nIeuXLlSi1CC7h7TZzYEJDCz4apcS0KgSA0TBAgKzJAuIUgIRsCnL1FoaAtTaILmVlYAKUv4Pl5WXHg91w/6FIFOCEhgdkJN11SGQNEaJqBwaGEsjbrgEXrkyJFKazsISLRmxiQxdc6bN89s2LDB/P73vzc9PT2luzcQ3MzpVBKBbglIYHZLUNeLwDgEWNcRxxMe3FUNccd4JKuwICQxs+Kwg9CsSsLzmLFkNH4lEeiUgARmp+R0nQjEJICwIVVxoWI33orJFc2SQAz8EaKuSonxUwnMKvVYOesqL9ly9otqlRKBvNfDDFYbcyVzCzEH4jFbBU9NhCKCBRMrk/4JdI5wvHjxYnPBZdpShPNOkG3SbTRLJ/STXqvzRcAR0DxMR0KfXhIoah4m5kse0C7EHUKIYOpMqeg2ek3aHYUGTF1ZqJlxVmKwYnKNG7M17fpklR/TWxCcAwMDWRWhfD0nIA3T8w6ue/OK8JJFS2OsLzi/Eq0MQenMs2XpF4QkghyByZxFBDwCxTdhCW/6oGz8y3IfqB7xCEhgxuOksypKIG8vWRaNZpI+gifoYOLmI545c6YwkoxHoumuXbvWCkQEJeHpnLnVacOFVTDjghlDpv0SmhmD9jh7CUyPO1dNy48AD2FMfvfu3bPBw5lGglkWk7DzJmUCPxPo80wISerx7LPP2vmfCIwPP/zQsB8BuXnzZi+1ySjGeCnjsOT6I+oc7ROBdgTkJduOjo6JQAwCRLRxK4245bwuX75sH84IUh7QRSxijJMRwpLPvr4+G1igLMEEYmDN5BTMspjMlUSgEwLSMDuhpmsqQyBrL1k0yfnz51vh2Gg0zNSpU602x6ojCCfMsiwG7RLrZDK+mXbCtIpgZsFkxiIxt/J979691tGI7boLS5jzQjPe6jFp943y84eANEx/+lItiSDw+PHjiL3d70JLQSihvbF+oxNGH3zwgZ1vSXQfEhpm0PFo2bJlZv369VaQdusti1l1165dVgBTBiHgEJjBUHzdt9S/HJgSQ/91y98/MmrReAQkMMcjpOOVJpCFhomG8sknn5jly5fbMHdOWGKaZYyQuZc490ycONFO1Zg1a1aTIRom56HxBQOyN09os4GAREB/88035uHDh+bRo0dWUDMFBE02j9U/2lSvMocwy+IdzMuFkggkISCBmYSWzq0cgbQ1TAQimuX+/fvHPHAxvaK1uHEytEtWJSF2rEuTJ0+2gpKAAEkEJtM+GI8kaTzS0ezsk/7BGUsCszN+db5KY5h17n21PREBVrxAWOJdGvWwZWwS7RKNk4cywpUUdvhxjkEI1FYJkyGmVbxb+SMv4tAS+EDjka2oxdvvXmhc/8S7SmeJgDHSMHUXeE0gDZMsZlCEFFNGEFo8cKMSAnDRokX2EM5APJDRRMNjZQhUzLnnz5+3whXT6u3bt625lTUlKYfEdZhbKY+/qoWji2JUhn1whT9ae3CubBnqpjqUm4AEZrn7R7XrkkC3JlmEIIEIEFholmHhF6wegpBxSxIPYybKE2YuKu3cudOu0UjM1kOHDlnnHc7DeQctlShBStkRwEKAtYDxZL2IZMfZt5xlkvWtR9We1Ag4T1iEV5xluRCQOAQx1ojQayUsyRdBjOMJ5lY02KC5VabC1LqwZUZolrzgMF9WSQTiEpCGGZeUzqskgU5NssePHzdnz561JtFWgi8KCEKTcHzPP/988zCrlVy4cMEKSeZLovWiSaK14vizbt26ppaDRy3HGA9NUm6zMG3EJsBYMi8rMsvGRlb7EyUwa38L+A2gE5MsD9F//Md/tNoiWkiSFAxajiZJXmicaJTkxUPaaTecy3HnBEQ5eNGinWLSlcBMQj75uXAnQpOSCMQlIJNsXFI6z3sCjGcxruWCpycVlgByczHJg9iyPJBx/MG7FQHIQsbO25bpIQhGrgkmYs5ilkXgKmVHgPHoMPvsSlPOPhCQhulDL6oNLQnENcni1YrzDWZSPFPjBgHggcs4GJoi2251FMysaIosvnznzh1rekWzJNIPD+qFCxdabZLyCMiOKZY8CEqAxyzCFe9ZpWwJoOXT93H7O9vaKPeyE5DALHsPqX5dEYhjkkVQoVlu2bLF4L0aJzlBibbI9c7cGhyPdPkwHsmD2QUqcEKV8UzMgu+++64VqmiUTHdg/mU7b1yXrz67J8ALy8GDB7WodPcoa5HDhAYRo5VEwFMCrP2IA0+rtG3bNqvhIaR4eLZKCEjmTRISj4g+OOagBaJFvvLKK83pJFHXY55lzUmXMNMSAWhkZMQ6+zgTsISkI5TfJ/NeX331VatldmKCz6+m9S2J3x4vNbyMBn0EiiCiMcwiqKvM3Ai00jAZI5wwYYIhSDpTQaKEJRofDjtM/Zg+fbr1mkWwMgWEa5jLx3Vu7mVUoxijRBCePn3aCkm+IyzxpnXz/8hD45VR9LLf9+KLL9qHMJYCpXIS4HfCb4jfYNFOWjLJlvMeUa0yJEAQc354/AhbeaJiZkVY8olAw3SKudQJubjV43rGMV05bo5lUEBTD2fWjZuvzkuPAGPL8FcqLwF+f/xO+N2632OnteX3H9RU8WAnxbIwYJJVEgFfCaxevXpU086dO9eYM2dOY3BwsLl/ZGSkMTQ01Ni+fXujt7fX/rHNvj/96U/N8zrZmDlzZuNf/uVfmpdu3LixsXv37uZ3Nr788svGlClTGtRDKX8C9+/fb0yaNKnxhz/8If/CVWIiAvRVf3+//b3s2LEj0bX8vvjdL1u2rLFkyRJ7LXmw7b6Pl6EZ7wQdF4EqEwgKTIQVwpIfDj+8EydOWOGIsOIHExSiabWZ8hDSJPKnDlGJOvAgUCqGAP2yf//+YgpXqR0RoL+MMfY3leRl8+LFi/Y6fvPDw8OJypZJtlO9XtdVigBmF8yhhLnDaceZW5kXOTAwYIMJYPJJO2E+wgTEGCXjZDguRCXOk1kwikw++2SWzYdzmqVgpuW3jF9AEjNt8HceNM3GqZu8ZONQ0jmVJeACbDNXkrFExiAZi2RMMa+5d3jXMh+UH3WrMVCEKoEOfv/73xscUZTyJ/Daa6/Zhb/zL1kldkuAqVmHDx+2vgZ4vhMxq1Xi5RVHPn6POPAlSYkEJoUoiUBZCbg3R4SPm+tIXd1+3NN5oyyrRyr1YrpKK6FaVu6+1Iv7Q+yr2Zv8xp3zDi/GBKNolViqjxdoLD7Dw8OjHIBaXeP2JzLJErJLSQTKSABBg5DkR8MqICR+RDwEMbfxx3ZQkJatHdQds6zzqC1b/VQfESgjAX7nCEjmapL4rbdLvJhyDgKTbcy6eMjG8ZJNpGG2q4SOiUDeBBCACBjeFnlrRBhiZmGc0Jk/Fy9ebL799lsb7o55k2VPq1atMidPnix7NVU/ESicAJGyCCTCCzIm2Pfff7857BKuHObaDz74wL6M/sM//IOdO03AEIQkPgxxhKXNM5GLkE4WgRIQOHXqVIPpGniW4t2I92krL7l169bZGnMuXrFlT1u2bMnEW7fs7Vb9RCAuATe1hKlAAwMDcS9L5bxEJtmw1NZ3EciaAOORmEwISn7v3j1bHOZXPOQwq6BJtkvOBHvq1ClrsmGFkKSece3yT/sYGjEevNRX42lp01V+VSWANQlLEp7ubGOG/dd//VeDo1aeSQIzT9oqKxEBTC27du2y4wwIOgKjjycgWxVAzFd+bOSHCaasad68eebv/u7v7EtCp20ta9tULxHohAACktV9GH5hJSHCShaVNIZZFHmVG0mAOVVoWCSEJLFbY48vROTICiHOCYgf3i9/+Uurva1YsSLi7HLscmMrzBlVEgERMIYVf7AooVkWmSQwi6Rf87Lv3r1rVwrB5MoAPoHSMbfyo2COJAKzW7PkmjVrbLAChxotE6GMlln0j8/VKfyJA9P69evNH/7wB9PT0xM+rO8iIAIFEZBJtiDwdS4WV27MK2iSTKVAcO3fv99kofUhgIOJt1QEEi7lRZp2gnUKb8MBrZqFpbNgEi7Pt+9YErp90fKNidqTDgEt75UOR+UyDgGEJGYVNEci2iAsEVj37983N2/ezEwwOKefYPUYC0VgOtNv8FhZtnFq4qVCKRkB7jGWTwsmXsrSXBaKewezOVGkgn3E/uD3YB207QcBaZh+9GPpWoF3K+bPO3fuWHMrJlc0SQQBDjguTF0RFaceeKHiAHTmzJncPe3itJngBRs2bIhzqs4xxjqG8SKEZs64tUssEM38O7ysu00MIfzmN7+x2WDSRzhS1u9+9zszf/588/rrr5tDhw7ZF0OcU6Tldku8hNenMjlFmYjAzwSYD8ncKJbJynIVkLjAg6uVhK+5efNm7GV9wtfm8b1d3fMov0plcM+1WqKJ+besGtNtOnLkiF3lgvvGJeYBu7m+bh/3vVuhxu3Tpx8EZJIt4UtM1aqEuRVHGsxUrIqOWQzPVMyt586d63gqSBocwmOYwTzRRvjDlFbGhBYOV6XxCXC/tZoulJZzFxYTd8+4GjEmzvzAYKIe6rcgEX+2ZZL1py9zawkCkhU4bty40QxkjmDCjMjcwTS8W9NqTNQYZjDvLVu2WI9UHEW2b99uQ2YFjxe5TRCDzZs3W65pPfSLbE9WZRNHFNM/wqxVon+7TYyFhsvgpYbhB/5cQAyEKCtm8DtRv3VLvVzXS2CWqz9KXRunSfL2zDYPBsYkyyQgwwDbaZicixMSGgHaMR6pZQoWQBADptyg2Sgge7hnn3zHyWc8wZTGeCL3fHhJOJcvAtsJTOrCFCn125M+8mVLJllfejKDdvDWfOnSJethyMMAJxTesC9evGj+53/+x85vRGi6h0YGVeg6y/E0TApASOKpi1ZQpgRXPIkJGv/o0aMyVa1UdeEenTp1auZ1QgiGBTNClHss/GJGfaiXkl8EJDD96s+uW4PnH9MtWPv05Zdfti76ePwhPFlFA00H4Tlx4sSuyypTBrwI8ELAlJcyJZbUQ3BeuHChTNUqVV0QWCzQnXVCCGKWDSYEJsLSaZfuGL+POC9r7nx9VoNA9ndZNTiolsZYR52tW7daFmhdaDdlMlFm3UlMd+GBiNNSmUyg9AEvLErFEkAoIiCDie9onWHNM3iOtv0hIA3Tn76M3RLMezyAiXhD6Di0KiZhY3IaGhoqhXdr7MaMc2ISzQMtk/ZjAoVNWRJRkBgPU4omwBi6W8km+oz2exF6vCiNZ/ZetmyZ7Yeglsl9wrBEOJEn9VLyjIAfs2PUijgEhoaGGsztY54Y60nu3r27wdpyPqdO5zIyj7RMafny5WWqTunqwj0dlZgX3N/fb+/54PxJdy73vzHG/nHeeMn9fsiLOZhcOzw8POayadOmjdmnHdUnIA3TsxegcHNwuUeLZEwSLRLz3vDwsHVyYXxMpqQwsZ++o2XiOVuWhHcmpmKlaALc11Fh6fAyxnKAJzRaelA7JCfuf4Ye8PYOm1ujSmIFGeYWs4oOmmWj0Rgzfkk/RWmdUflpX7UIaLWSavVX29oy14wHAnMkMVHxAJg8ebL948HAj5iHR53S2rVrzfHjxxM3mTBov/71r82qVatKEaLOBYznYV1mr+TEoFO6wAXLQDCGHXDiFIEwJYwe817jpFYB3nlBJW4tYfokNOOQrNg51VeS1QIIDA4O2tBgztxKqDDfza1xer5Tk6zLO42Qai6vbj6d6VAh11pTxKTaSX8TOo/r0vi9EJ6PoQ4lPwlIw6zYC06wumiQhHUjkIAzLfFWKzPrE0rh9TCfHIm3hent9OnT1gwX74rszsKkTkLLVIomgFkWLS8YgD36zCd7+R2l8ZvhXnFLsz3JXVs+EZDArEhvYgJiPIYHwq1bt6xHK1V/5plnzIwZM6z5pxNTVEWa33E1OzXJugLxJmYxZ0zZe/futeZtdyzvTybCr1y50nryqq9b06fPiuDD75LfopK/BDQPswJ9y7gkYyJoOiTinxKVJo234go0v6sqMlWmm8SDF0cPpt6gvRc5P5N5oji3EOy7yHp0wzOPa4sQlrRLwjKP3i22DAnMYvmPKR1NEi0SJwQi6+C0g4ckUWhwaJDDxxhkme/gxQTPYkyiRZu8mZN58OBBCczMe10FiMBYAhKYY5kUsofliVjQmPEUtAfc3F1YtEIq5Emh4RifnTaLFxW0fKbnEHe2KG9jXp4U9afTXtR1ItAdAQnM7vh1fTXapHOJJ/akmxPWdcbKwBJIM56ni9yCoxXaflFJVoaiyKvcuhNQ4IKc7wAEJPO0MO/Nnj3bCksexGgtLLiMZqmUHoEkofHGKxVBNTIyYr0w8YgsKhGiLYkXaFH1VLki4BsBeclm3KOYWJ25laIQjjxsizLpZdzc0mXf7bSSVg0iChDOQEX0IybZ6dOn23HVohxcWnHRfhHwmYA0zAx7l/lgLJXF2CSOI2iPCM8iHrIZNrOWWfPiw/zXIpITkixarCQCIpAfAWmYKbHGu5UHGGHYPvvsM+u8g+ns9ddft1MBNO6UEuiE2WSlYdLfCxcutC9C//Zv/5b7+qAIa8L3vfvuuwmJ6HQREIFOCUjD7JScMVYo4rDDHD1MZDycFy9ebAYHB+2Y5IcffmiYOydh2QXkLi9Ny0s2XA36lH7GcvDBBx+ED2f+HWvFlStXMi9HBYiACDwhIIH5hEWiLZx3EJA48JDwbiVkGQJSqTwE0vSSjWoVU3+0VmUUGe0TAf8IaFpJjD5lYdlvv/3WerS6hYVZxYJJ5IxHSoOMAbGgU9L0ko1qApF3iB/Kn7s3os7LYh9e1oyJy2M2C7rKUwTGEtAY5lgmdg/erZhbGSsiNB0PRrRIPpWqQyCrMcwwAaYJYXVgalBeifLw1mVNRiUREIHsCUhg/swYAck6kjdu3LBjk+xm/Avtcd68efZPmmT2N2TaJeQlMHmpIqDBrFmzbJzftNsRld+DBw/Mq6++aiNCaf5uFCHtE4F0CdReYDKn7dixY1aTRGgSK5Rxqd7eXpla073XCsktL4HpGkfoOqYT5ZWYtsQYKg5ISiIgAtkSqK3TD2M/eLdi0kI7YBI6Abb5xOwqbTLbGy+v3LPykm1Vf162uKfySpSHaTZPIZ1X21SOCJSNgPdOPzjs3Lt3r7mWJFokc+hcIAE0So1Llu22TK8+WXvJhmuK8w/LsDGlaPPmzeHDqX/nPl63bp1d8gvtVkkERCA7At4LzI8//tiOLSEoFZYuuxuprDnnrWEiwLBSoGXmITDhzioqjJ8qiYAIZEvAO5MsgpFYrSzDxB/jO8yPxHsRt3+Fpcv2hipb7nlrmK79aH15mWa5pxlWUBIBEciWQOU1TMKDXbhwwVy/ft2aXsHFW/6OHTusqVXjkdneQMo9mgAhEXlZY44kL3Dck1kmhOalS5cUOCNLyMq79gQqKzDRJHF2wEuQt+vly5db05Q0yNrf06UA0NPTYy0aOJZh2cg6uAAvhnh7K9JUKbpflfCUQKWmlSAkeWtnzIZA5whHggnguKMkAlEEEFR4RBeZEGJof1kmXhoRzgw9ZK3NZtkO5S0CZSZQaoHJHEk8DhGU/DnvVh4IPIQ0V7LMt1Y56pb3PMyoVuOQg0BjmMAtzRV1Xjf73OopzrGtm7x0rQiIQDSBUppk+fFjXuJBg6DE3KSwdNEdqL3tCeTtJRtVmy1bttjVbLiX8aDNIjFvmGg/DFHwW9E84iwoK8+6EyiFlywPEuK2og1gVlq5cqUNS/fll19aExNerporWfdbtbP2F+UlG6wtwougGLwI7tu3L3go1W0EJtYXQjwqiYAIpE+gMIGJiQoBydQPhCTfBwYG7DqSCEjGnnCcUBIBHwggNHH+OXv2bKZTQNBgWcBcSQREIH0CuQtMtElMrazugAPP6tWrrZkKs5WcFdLvYOVYLgKMMWYZZABHOH5jSiIgAukTyHwMkx/v8ePH7Vsv8S5x1mEKyMWLF83UqVPNxIkT02+VchSBnwlkvR5mUtDEfuVlkaAG/AbSHmvkpZNwkPzu9AKatHd0vgi0J5CJhunmR+IRyINh8uTJNtoOXq+YjDC3ckzCsn3n6Gj3BB4/ftx9JinmgIBkiglj8jjoZJEWL16cqRabRZ2VpwhUgUDqApMFl3mD5pOHghOQVYChOvpHoAxeslFU3SojUce63YdZluEOJREQgXQJdCwwMfvgqIN366pVq6zGiKDEBHbz5s1m7FZ5t6bbYcotGYEyeMlG1Rihxu+G3wzes2kmNEz320wzX+UlAnUnkEhgMgbJH96tv/rVr+yPnR/m4cOHDeZWvFs3bNigsZO631VqfywC27Zts9Gq0nYCYggEDTYrk2+sxukkEfCQQCKnn/Xr11sECEnFbvXwblCTcifggg2kXTDhIhGYcv5Jm6zyqzOBRALTRSnBnKQkAlUgUDYv2TAzhiwQbiw8zTzNtBL5MmXr5MmT1uqTVr7KRwTqTCCRSRZBKWFZ59ulem0vm5dsFEHMp4y1EsAjzfFMQuRduXIlqkjtEwER6IBAIoHZQf66RAQKJVBWL9kwFIQmQx3Xrl0LH+r4Oy+3+BYoiYAIpENAAjMdjsqlpATK6iUbxoVZFu9yHIHSTAQHwVFPSQREoHsCEpjdM1QOIpAKATRCvMwxzaaVmGKya9eutLJTPiJQawISmLXufv8bX3ann3APLFu2zAb8YOpWGjFhCUVJEAOZZsOk9V0EkhOQwEzOTFdUiEAVnH6COFmhB2/0tAIPOEe9y5cvB4vRtgiIQAcEJDA7gKZLRCBrAoODg+bAgQOpFJNlGL5UKqhMRKAiBCQwK9JRqmZnBKpmkg22ctOmTXahArTNbhJBRrrNo5vyda0I+EJAAtOXnlQ7IglUzSQbbMTmzZsNptS33noruDvxNiukML8zjTHRxIXrAhHwiIAEpkedqab4RwDTLNrh0aNHu2oc01ZYQUhJBESgcwISmJ2z05UVIFBlkyx4WQT6yJEj5vTp0115us6ePdsKXWmZFbhpVcXSEpDALG3XqGJpEKiySda1n6khvb29xi1+4PYn+WQ+JlGEtE5mEmo6VwRGE5DAHM1D30SglAScp2s345nkoSW/Stm9qlRFCEhgVqSjVM3OCFTdJOtajePOqVOnbJg7Fm3vJK1bt846/ly/fr2Ty3WNCNSegARm7W8BvwH4YJJ1PYRpliXAOl1w+qWXXrJRhLSCiSOqTxFIRkACMxkvnS0ChRLACQhNce7cuR0tBUYUobNnzxbaBhUuAlUlIIFZ1Z5TvWMRqMryXrEa8/NJb7/9ttUUOwmqjsB98OBBkuJ0rgiIwM8EJDB1K3hNoCrLeyXtBBx4Ll26lPQyez5CU5F/OkKni2pOQAKz5jeA7833xekn3E8EVV+1apVZunRpYtMsK6KgpSqJgAgkIyCBmYyXzq4YAZ+cfsLoWWwabXH69OmJhCaClkAImpMZJqrvItCegARmez46KgKlJrB79247VeTrr7+OXU+mqKChSmDGRqYTRcASkMDUjSACFSaA4BseHrYBCQiwHjft37/fTk+RA1BcYjpPBIyRwNRd4DUBH71kwx323HPPmY0bN5qXX345fKjl9xUrVlhPW00xaYlIB0RgDAEJzDFItMMnAr56yYb76I033rBm1hdeeCF8qOX3gYGBjj1tW2aqAyLgMQEJTI87V02rFwEEYF9fn9m3b1+shhPQ/datW7HO1UkiIAIyyeoeEAFvCDCeybqZhw8fjtUmnH8mTpyohaVj0dJJIiCBqXvAcwK+zsNs12179+41jGtevXq13Wn2GAtLa07muJh0gghYAjLJ6kbwmoDP8zBbddxrr71mDhw4YJ2Axovos3r1anPs2DFF/mkFU/tFIEBAAjMAQ5v+EaiDl2xUryEI48y1JPABf5qTGUVR+0RgNAEJzNE89M0zAnXxko3qtps3bxrWzhxPyyT4AWbZJPM4o8rTPhHwnYAEpu89rPbVmgDCkKkm7YQmczgZ85SWWetbRY2PQUACMwYknSICVSXA1BFMruOtbILzz507d6raTNVbBHIhIIGZC2YVUhSBOnrJBlkzdeT+/fvmo48+Mt99913w0KhtNNHz58+P2qcvIiACowlIYI7moW+eEaijl2xUFyIQWdWkVZB2tFCxiiKnfSLwhIAE5hMW2vKQQF29ZMNdOWfOHGua3bNnT/hQ8ztCs5VAbZ6kDRGoMQEJzBp3fh2aXmcv2WD/YpplVRPMsszRjEoKYhBFRftE4AkBCcwnLLQlAl4TQGji/NNqqsny5cvN5cuX7XGvQahxItAhAQnMDsHpsmoQqLvTT1Qvbdu2zaxZs2bMNBIEKlNMMNs+evQo6lLtE4FaE5DArHX3+994ObKM7eNVq1YZzK9Lly4dE3i9v7/fjmPevn177IXaIwI1JyCBWfMbQM2vJwG8ZtEmw+OZhNMjrN548zbrSU2trjsBCcy6qev2zwAADZFJREFU3wGet18m2dYdjNC8du3aGOG4ZcsWecu2xqYjNSYggVnjzq9D02WSbd3LTCNBm1y5cuUo0+yCBQusIG19pY6IQD0JSGDWs9/VahGwBBCYjGcSbzYYfB1h2i7+rPCJQB0JSGDWsddr1GaZZMfv7P3791tN89ChQ82TFy1aNGZ8s3lQGyJQUwITGo1Go6ZtV7NrQIDpEydOnKhBS7tv4ty5c83g4KDN6OrVq3YB6pGREcN0EyUREAFjpGHqLhABEbAEcPbBNEt66aWX7ALUBDJQEgER+ImABKbuBK8JyCQbv3tXrFhh+vr6bFCDu3fvGgToV199FT8DnSkCnhOQwPS8g+vePHnJxr8DcPQ5evSodfb57LPPzNtvv22uXLkSPwOdKQKeE5DA9LyD1TwRSEqAcUwXb3by5MmjvGeT5qXzRcAnAhKYPvWm2jKGgJb3GoMk1g48ZxnPnDVrltm6dWusa3SSCPhO4CnfG6j21ZuAlvfqrP+d0w+RgDQfszOGuso/AtIw/etTtShAQE4/ARgJN2/evGmFJYtKM7apJAJ1JyCBWfc7wPP2y+mnuw4eGBiwGcgs2x1HXe0HAQlMP/pRrRCBTAj09vba+Zjff//9qHizmRSmTEWg5AQkMEveQaqeCBRJgCg/Q0NDNnTeyZMni6yKyhaBwglIYBbeBapAlgTkJds9XYQm4QVl3u6epXKoNgEJzGr3n2o/DgF5yY4DKMFhFpdWEoE6E5DArHPv16Dt8pJNr5NZBkxJBOpMQAKzzr1fg7bLjFiDTlYTRSAnAhKYOYFWMSIgAiIgAtUmIIFZ7f5T7UVABERABHIiIIGZE2gVUwwBeckWw12lioCPBCQwfexVtalJQF6yTRTaEAER6JKABGaXAHV5uQlIwyx3/6h2IlAlAhKYVeot1TUxAWmYY5F99913Zs+ePWMPhPawJiaLSHebKI+8+CPEXrv0+eefm6VLl5q5c+eaS5cutTu10sfgAI+PPvrIwKddIvj9G2+8YbkEz+V6pXwJTGg0Go18i1RpIpAfgTVr1tgoNfmVWO6SePh+9dVXZvHixYYlvNolHuZXr141Bw4caHda22M81I8fP24ePnxoBQMLUr///vsmPKeTco4dO2amTp1q68aSYh9++KE5f/582/yreJC28sKC0Lxz54558OCB2bRpk9m4caMhqpJLt27dMocPHzbPPfeceeWVV8yVK1fMN998Y/dxzvXr180nn3xipkyZYq+fOHGiu1SfWRFAYCqJgK8EVq9e7WvTErfr/v37jZkzZzYuXryY+NpOLnDlDQ0N2ctv3rzZmDZtmq1DML+RkZHGnDlzxuynrlzjW+KePHLkSLNZfX19KC2NEydONPexsXHjxsbu3bub++DZ29vb/O42pkyZ0jh37pz7qs8MCcgkm9WbiPItBQGNYT7pBkydaCMLFix4sjPDrS+++MKwPJjTZAmth4kX7ZFjLqFZovmeO3fO7bKfP/7446jvPnxBq5wzZ441sbr29Pf3280gE8fIHeMEruUvnIjz240VIJyfvrcm8FTrQzoiAtUnoDHMn/oQMyxC6ciRI6M6FRPfjRs3rCDlIY1AXb16tTWfYk7FRIipEOHF92vXrpl169bZvMiP42+++aZ58cUXR+XLF8yu5BdMzzzzjOElhmXDXGJxagRDMFYtZktMleHr3TVZfN69e9eaOGfMmGFNn67trg4INAQW9YcDbVixYkWzKrSDNnAOx4PCzp1E++EZTC5/x4Tr33rrLfO3f/u3wdMsc3du8EBfX5/ZvHmzfREJMgyeo+10CEjDTIejcikpAWmYP3XMhQsX7EZYu9y3b58VbAg3WDlHEh74u3btan5nrI0xTcbe2O+EJd8ZR4tKUQ/327dvW+2KcTkS+fAXFi6nT582PT09uQrM3/72t1bQrVq1yvCi9c477xickFzauXOnZYWQ5AVi5cqVTT6uHfDleFBbdNfzyThjmAts2YdjD4k+4PrXXnsteKndHyUQEeyMDbu+G3WRvqRKQBpmqjiVWdkISMP8qUfQ2HjYhh/WaJc8aNEq+XMJAYoQcw9+rsWcevnyZathOs0KrZQ84iSEDPUICiGExKxZswzamesrPEEPHjxoNdc4+aZxDlod9aJ+JNrOPpyQSAhCHKUcv+3bt1snJZzKWC8UTghN6s7LwO7du+11cf6xzih5kDdaNZx5eUH7J08S+dJPe/fujczy+eeftwzDLx6RJ2tnxwQkMDtGpwtFoDoEEEY8kN0D39Uc78z169dbEyCmVgTXeAmNJpjiBLhH+DBV5NSpU81LEbZosjt27Bg1fQUhRFq7dm3z3CQbaIZO0ERdh+kzPOaHwAuyYZvxVxJ58aLgXhLYh6bICwYatjO/wm769Ok2n2A7o+rg9lHXZcuWWSHLPjxh6SsE7rZt29xpzWlAaL9RCQ0TTVUpWwISmNnyVe4iUGoCaJL8ITDQYBAKQU0zrcpjzkVLdQ5A5IsQJQXNjGihTtgF9yepx3jXubHCYJ7UxWm4wf1su3qG9wcFLMfQ7nj54HzM1mEnpvD1tBPmFy9ebB5yZYXb4CwE06ZNa56rjfwJSGDmz1wl5khA62H+BBtnE8x6PJCDD3pMgDzYEVRoO2h3OAa58bQ0ugphiZAKmgupy6NHj2z2ONm4xHgodUSIhIWGO2e8z7BTzXjnc9w5zhAsITjOi7kUDRBmmEiDyQk3TMqYc3nRgBt5wJExT/6i0pkzZ8zZs2etKdZp7PSBy5P6uAQrNM/BwcFR8zTdcT7px055BfPRdnsCcvppz0dHK04gjrmw4k2MVf358+dbkx1m0GB677337FfnEcqXoBYYPLeTbYQwAhMnHjw/+cM0iybmBKXrI4TFoUOHzP3790cJLVcuQohrs0i0mTHKhQsXWiFHnZ999lnrTAMbXioQ5k77RbhhjkU4z54925psmR5Dco4/zrEpXF+0+d/85jdW+G3dutUyoW2Yc53Qc9ou57L/d7/73agXnXCeOFMFX0jCx/U9JQIZzvFU1iJQOIF169YVXocyVODUqVN2cnx4gjsBAwYGBhpMiueT7ySCCSxZssQGE3DBA3bs2GHz6O/vt8fZz/n8cX04kR8T8qP+3IR8ghOQL3kxoZ+8oxL1cfm4+kSd180+6uTK4JO6Ua5LBA2ACeUTeCAYWIF6cwwO/BGMIIoJ+7guWI7bZv/w8LA9Rl0c3zjBN4J1cfXVZ/oEFBovpRcPZVNOAry5M7G77gnz59/8zd/YcTpMey4x9xDtBLMfHqFooiS0PWce5DtjZ0GnEkyUweN4dYY1qrA268rk02lSbKNFcT37wnkEr2EcDxMpcz6zGGelLDjgxENdgnXkGAwxjcIBMyyapUuOl/OyRcsMmr7deWimQY5uP58wRpvlHDRZNO958+a1NMO6azEBMz0INkrZEpDAzJavci+YgATm6A6Ax/Lly1MdoxxdQrbfMPEmmbKRbW2Kz52XEszUvASFBXzxtfOvBhrD9K9P1SIRaEmAcS43DtfypJIecM5DJa1eIdViHBVtW8IyH/zSMPPhrFIKIsBcPlbLUHpCgFUw8PgMzit8crS8W5gqnUdpeWuZX81YyQXTcDszdn61qUdJmlZSj36ubSudB2ZtAUQ0HO9U56Eacbi0uyQsR3cNXr1K+RKQSTZf3iotZwKah5kzcBUnAh4TkMD0uHPVNGMkMHUXiIAIpEVAAjMtksqnlATcBPBSVk6VEgERqBQBCcxKdZcqm5SANMykxHS+CIhAKwISmK3IaL8XBOT040U3qhEiUAoCEpil6AZVQgREQAREoOwEJDDL3kOqnwiIgAiIQCkISGCWohtUiawIEKNUSQREQATSICCBmQZF5VFaAvKSLW3XqGIiUDkCEpiV6zJVOAkBeckmoaVzRUAE2hGQwGxHR8cqT0BespXvQjVABEpDQAKzNF2hioiACIiACJSZgARmmXtHdRMBERABESgNAQnM0nSFKpIFAXnJZkFVeYpAPQlIYNaz32vTannJ1qar1VARyJyABGbmiFVAkQSkYRZJX2WLgF8EJDD96k+1JkRAGmYIiL6KgAh0TEACs2N0ulAEREAERKBOBCQw69TbaqsIiIAIiEDHBCQwO0anC6tAQGOYVegl1VEEqkFAArMa/aRadkhAY5gdgtNlIiACYwhIYI5Boh0+EZCG6VNvqi0iUCwBCcxi+av0jAlIw8wYsLIXgRoRkMCsUWerqSIgAiIgAp0TkMDsnJ2uFAEREAERqBEBCcwadXYdm6r1MOvY62qzCGRDQAIzG67KtSQEtB5mSTpC1RABDwhIYHrQiWpCawLykm3NRkdEQASSEZDATMZLZ1eMgLxkK9Zhqq4IlJiABGaJO0dVEwEREAERKA8BCczy9IVqIgIiIAIiUGICEpgl7hxVrXsC8pLtnqFyEAER+ImABKbuBK8JyEvW6+5V40QgVwISmLniVmF5E5CXbN7EVZ4I+EtAAtPfvlXLREAEREAEUiQwodFoNFLMT1mJgAiIgAiIgJcEpGF62a1qlAiIgAiIQNoEJDDTJqr8REAEREAEvCQggellt6pRIiACIiACaROQwEybqPITAREQARHwkoAEppfdqkaJgAiIgAikTeD/A0w6K/s2rTYkAAAAAElFTkSuQmCC"> <em><strong>(M1)</strong></em></p>
<p>using the transformation of the unit square:</p>
<p>\(\left( \begin{gathered}<br> 1 \hfill \\<br> 0 \hfill \\ <br>\end{gathered} \right) \to \left( {\begin{array}{*{20}{c}}<br> {{\text{cos}}\,2\theta } \\ <br> {{\text{sin}}\,2\theta } <br>\end{array}} \right)\) and \(\left( \begin{gathered}<br>0 \hfill \\<br>1 \hfill \\ <br>\end{gathered} \right) \to \left( {\begin{array}{*{20}{c}}<br> {{\text{sin}}\,2\theta } \\ <br> { - {\text{cos}}\,2\theta } <br>\end{array}} \right)\) <em><strong>(M1)</strong></em></p>
<p>hence the matrix <em><strong>P</strong></em> is \(\left( {\begin{array}{*{20}{c}}<br> {{\text{cos}}\,2\theta } \\ <br> {{\text{sin}}\,2\theta } <br>\end{array}\,\,\,\,\begin{array}{*{20}{c}}<br> {{\text{sin}}\,2\theta } \\ <br> { - {\text{cos}}\,2\theta } <br>\end{array}} \right)\) <em><strong>A1</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>using the transformation of the unit square:</p>
<p>\(\left( \begin{gathered}<br> 1 \hfill \\<br> 0 \hfill \\ <br>\end{gathered} \right) \to \left( {\begin{array}{*{20}{c}}<br> {{\text{cos}}\,\theta } \\ <br> {{\text{sin}}\,\theta } <br>\end{array}} \right)\) and \(\left( \begin{gathered}<br>0 \hfill \\<br>1 \hfill \\ <br>\end{gathered} \right) \to \left( {\begin{array}{*{20}{c}}<br> { - {\text{sin}}\,\theta } \\ <br> {{\text{cos}}\,\theta } <br>\end{array}} \right)\) <em><strong>(M1)</strong></em></p>
<p>hence the matrix <em><strong>Q</strong></em> is \(\left( {\begin{array}{*{20}{c}}<br> {{\text{cos}}\,\theta } \\ <br> {{\text{sin}}\,\theta } <br>\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}<br> { - {\text{sin}}\,\theta } \\ <br> {{\text{cos}}\,\theta } <br>\end{array}} \right)\) <em><strong> A1</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><em><strong>PQ</strong></em> = \(\left( {\begin{array}{*{20}{c}}<br> {{\text{cos}}\,\theta \,{\text{cos}}\,2\theta + {\text{sin}}\,\theta \,{\text{sin}}\,2\theta } \\ <br> { - {\text{cos}}\,2\theta \,{\text{sin}}\,\theta + \,{\text{sin}}\,2\theta \,{\text{cos}}\,\theta } <br>\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}<br> {{\text{cos}}\,\theta \,{\text{sin}}\,2\theta \, - {\text{sin}}\,\theta \,{\text{cos}}\,2\theta } \\ <br> { - {\text{sin}}\,\theta \,{\text{sin}}\,2\theta - {\text{cos}}\,\theta \,{\text{cos}}\,2\theta \,} <br>\end{array}} \right)\) <em><strong>M1A1</strong></em></p>
<p>\( = \left( {\begin{array}{*{20}{c}}<br> {{\text{cos}}\,\left( {2\theta - \theta } \right)} \\ <br> {{\text{sin}}\,\left( {2\theta - \theta } \right)} <br>\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}<br> {{\text{sin}}\,\left( {2\theta - \theta } \right)} \\ <br> { - {\text{cos}}\,\left( {2\theta - \theta } \right)} <br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br> {{\text{cos}}\,\theta } \\ <br> {{\text{sin}}\,\theta } <br>\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}<br> {{\text{sin}}\,\theta } \\ <br> { - {\text{cos}}\,\theta } <br>\end{array}} \right)\) <em><strong>M1A1</strong></em></p>
<p>this is a reflection in the line \(y = \left( {{\text{tan}}\,\frac{1}{2}\theta } \right)x\) <em><strong>A1</strong></em></p>
<p><em><strong>[5 marks]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong><em>Q </em></strong><em><sup>T</sup></em><strong><em>Q</em></strong> = \(\left( {\begin{array}{*{20}{c}}<br> {{\text{cos}}\,\theta } \\ <br> { - {\text{sin}}\,\theta } <br>\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}<br> {{\text{sin}}\,\theta } \\ <br> {{\text{cos}}\,\theta } <br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br> {{\text{cos}}\,\theta } \\ <br> {{\text{sin}}\,\theta } <br>\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}<br> { - {\text{sin}}\,\theta } \\ <br> {{\text{cos}}\,\theta } <br>\end{array}} \right)\)</p>
<p>\( = \left( {\begin{array}{*{20}{c}}<br> {{\text{co}}{{\text{s}}^2}\,\theta + {\text{si}}{{\text{n}}^2}\,\theta } \\ <br> { - {\text{sin}}\,\theta \,{\text{cos}}\,\theta + {\text{cos}}\,\theta \,{\text{sin}}\,\theta } <br>\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}<br> { - {\text{cos}}\,\theta \,{\text{sin}}\,\theta + {\text{cos}}\,\theta \,{\text{sin}}\,\theta } \\ <br> {{\text{si}}{{\text{n}}^2}\,\theta + {\text{co}}{{\text{s}}^2}\,\theta } <br>\end{array}} \right)\) <em><strong>M1A1</strong></em></p>
<p>\( = \left( {\begin{array}{*{20}{c}}<br> 1&0 \\ <br> 0&1 <br>\end{array}} \right)\) <em><strong>AG</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p>The transformations <em>T</em><sub>1</sub>, <em>T</em><sub>2</sub>, <em>T</em><sub>3</sub>, <em>T</em><sub>4</sub>, in the plane are defined as follows:</p>
<p><em>T</em><sub>1</sub> : A rotation of 360° about the origin<br><em>T</em><sub>2</sub> : An anticlockwise rotation of 270° about the origin<br><em>T</em><sub>3</sub> : A rotation of 180° about the origin<br><em>T</em><sub>4</sub> : An anticlockwise rotation of 90° about the origin.</p>
</div>
<div class="specification">
<p>The transformation <em>T</em><sub>5</sub> is defined as a reflection in the \(x\)-axis.</p>
</div>
<div class="specification">
<p>The transformation <em>T</em> is defined as the composition of <em>T</em><sub>3</sub> followed by <em>T</em><sub>5</sub> followed by <em>T</em><sub>4</sub>.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Copy and complete the following Cayley table for the transformations of <em>T</em><sub>1</sub>, <em>T</em><sub>2</sub>, <em>T</em><sub>3</sub>, <em>T</em><sub>4</sub>, under the operation of composition of transformations.</p>
<p><img src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAbgAAADgCAYAAABxYHOdAAAf/UlEQVR4Ae3df2wUZ57n8Y8NyeYCeBDij3QDwhc8MWRJDsWtRYq4W/Bu7FhcBAcKZFYKjMwf7MGcmEW4WWYj3dw5CiJYyUXKajJZ2ZsEToKbcYaZyRwBtjtkj4QDdcN6EgW3sS1n1kMPAk2w3RN5iF11qjZwRXebduMud1X53ZLlx/Xjqed5fcvPt7vrV5lpmqZ4IYAAAggg4DOBcp/1h+4ggAACCCCQFpiJg/MCZWVlzm+ELSCAAAIIyP6lJAluCnYIC9xKcnb4KdjstNgErs6FGVtnbHF1xtWq1bK1v/iK0q5BGQEEEEDANwIkON+Eko4ggAACCNgFSHB2DcoIIIAAAr4RIMH5JpR0BAEEEEDALkCCs2tQRgABBBDwjQAJzjehpCMIIIAAAnYBEpxdgzICCCCAgG8ESHC+CSUdQQABBBCwC5Dg7BqUEUAAAQR8I0CC800o6QgCCCCAgF2ABGfXoIwAAggg4BsBEpxvQklHEEAAAQTsAiQ4uwZlBBBAAAHfCJDgfBNKOoIAAgggYBcgwdk1KCOAAAII+EaABOebUBbekZF4i6rKytLPULKeo5T7Z5PauoYLr3xar5FSvGXNOJ425/o2dRnTGqrwzo/E1VJlM8y53wZV39YpaAvj9eN4QIIrbB/w0dLD6u24qEXhVkUSAzLNa4qEa6TAPkUGRtMPZx3tb1fj2me1quohH/V7Crpi9Kvj1CyFWyNKDI3KHIgoHJAC4YgGTFOm+Uf1t+/S2s1Pq4r/wIICYvR26NSisFojCQ2ZoxqI7FNANQpHrqX3WXO0T+2NDdq8qlLQFkLrz/GAJ3oXsg/4aVmjT2f7Nunw/nVaYI0Eg1GdOBSXntihhbPHhobyOfO0/PnHGYQLjLvRfUF92/9e+zcsVrkMDUb/SYeSAT1RHdTsdF0zNWfun+r55QzChdEOq/vsgLYfbtaGBQ9Kuq7oiZNK6lFVLxyTVfkszV1ep+W8KSuM1qfjQZlpmmZhEix9PwJuf0y90dWmhuqXpNaojjcu9cy7X7e7SsPqatui6m1Sa+I9NT7mnU/Drrc1OtXWUKttelmJ4416zCMf2VzvKskv44FHdon7SSmsM3GBYXWf+VAnFdSKyvmeSW4T718JlzT6dOboGSlQpcpHrE8dvIolYHR/qqMnkwqsqNQjjGTFYpX1pswv4wG7RRF3C+9WlVJ/olcK1Kk+NM+73XBjy1NXlPgsqcCLf6lQBf9uxQuRoVR/tz5TjV6sf1IVxauYmuSf8YD/OHZnaeRLXWi3jr9V3Tn+BktxBEYuX1D7XcffilMvtXytyxfO3n38DZTiCPhoPCDBFWeX8HQt6TPTegKq46y+IsfROjPtvHq0irP6iiyr9JmqCamOs3yLTmudqeqT8YAEV+y9w3P1jejq5+c5/uZI3K7p848vcPzNCdurX+hjjr85IOuv8YAE58Au4q0qb+jSudg9jr/9q97fWqWysjVqiae81bVSt3bwss6d6sl9/M1I6vzrWxW0LlQOblNb52CpW+uh7RsavBTTqfGOvxm/VfQH9ekL7YNb31Vniku+Jx7cfOOBVZOhVPx1ralqUXxk4jWXYkkSXCnUXbNNa0c9pOZX49IzIS3LeRLEIm149yO1b5nlmlZ7oyE3FH/7Nb2aXKJnVn476yQIo/sjHftWk7rMrxTbfVXbdv2Uu5pMNLCpmN5ufkdJhbRy2dyMtQwNnj6mjo1HZV30/ebMY/og8XXGMvyZW2Ai44GkVExv7Tmo07krcddU6zo4Xs4LSNaNFlz0GoiY4YCsayBtPzVmOHItRyN/Y7ZvWWsejA3lmFfaSa5zNUfNgcg+M3CXq0wF9pmRgdGcWKOJVrPuHvNzrjQFE91ne82MhGts++vYvhsIR8yBLI9RcyjRboa3vGHGhnK7Z60yRRPc52qa5oTHg6/M2Gv7zSM/+aG5ZMlBM/bNFKFNcDOZttzJxF3vN6auNRW1OnDF1IGp2+I02VK5Kmpf0RXzlYL6+0BDSNW37iBT0IrTauH5qj0Qk5l3px1R8v3vKbjxE2157S0FH+aLqry7yYTGA+sT3hG1L/6Ofrj4J/q7vJWWfgEiX/oY0IJpLXBDF3/xa63eUTt2y7RpbVGszs9UYMNbGr3yllb8yy599x1uvFwUWaNL7T8PaOd66xZ03nh5pZ3e0KSVCBQkcFPJ6BH981NN2l2TeSypoIpYOIdAeWClXlgXUs/v/8CTBXL4FDpp5OIHam5er4UzyvRAqEk9PU0KLXX3iSYkuEKjzPIIFEXAUKrzZzoy1KBdtQuk5EWd7uJMyqLQ3q7EuKpLsSFtfGqxOBZzG+X+f8+s2aPu9NMwTH0TO6glSw4q1rlHNS7GJcHdf7ynyZrWs822aON7v1JT6DkuFShK1G8qef5H2ln7gnavr9SMsjLNCO5Xr7hX5eR5rZtbbxp7Ft+MRp2o3qXvr54/+WqpwZMCPE1gisLmhTuITxFFUTeDa1E576oM27s4ivYHrkWjzKoo05ZPcFlETEAAAQQQ8IMACc4PUaQPCCCAAAJZAiS4LBImIIAAAgj4QYAE54co0gcEEEAAgSwBElwWCRMQQAABBPwgQILzQxTpAwIIIIBAlgAJLouECQgggAACfhAgwfkhivQBAQQQQCBLgASXRcIEBBBAAAE/CNx1JxPrKnBeCCCAAAIIeFXANK1HBo69sm6TaZ95eyF+T14g8xYyk6+RGiwBXJ3bD7B1xhZXZ1xvjwf22vmK0q5BGQEEEEDANwIkON+Eko4ggAACCNgFSHB2DcoIIIAAAr4RIMH5JpR0BAEEEEDALkCCs2tQRgABBBDwjQAJzjehpCMIIIAAAnYBEpxdgzICCCCAgG8ESHC+CSUdQQABBBCwC5Dg7BqUEUAAAQR8I0CC800o6QgCCCCAgF2ABGfXoIwAAggg4BsBEpxvQklHEEAAAQTsAiQ4uwZlBBBAAAHfCJDgfBNKOoIAAgggYBdwf4Ibiaulqiz9WBTrMRO5f4Kqb+uUYe8Z5bwCI/EWVY1rett6k9q6hvPWxQJ2gZTiLWvG2Vdvu5aprL5NXey0drj8ZcaD/Eb3uYQfxwPXJzijt0OnFoXVGkloyBzVQGSfAqpROHJN1rPrzNE+tTc2aPOqSrm+M/e54zmz2rB6Oy5qUbhVkcSATPOaIuEaKbBPkYHRtO1of7sa1z6rVVUPOdMEv9Zq9Kvj1CyFWyNKDI3KHIgoHJAC4YgGrH3W/KP623dp7eanVcVOW9BewHhQEFcBC/tzPMh64GkBIlOw6LC6zw5o++FmbVjwoKTrip44qaQeVfXC2WPbL5+lucvrtJxBuLB4GH0627dJh/ev0wJrkB2M6sShuPTEDi2cPTbqls+Zp+XPP84gXJisjO4L6tv+99q/YbHKZWgw+k86lAzoieqgxvbamZoz90/1/HLelBVGy3hQmFcBS/t0PCgzbY/wdv2TZo1OtTXUapteVuJ4ox7z0Ltft9saXW1qqH5Jao3qeONSz3wadrurNKyuti2q3ia1Jt5T42Pe+TTseluPjgeud5Xkl/HAQylCMro/1dGTSQVWVOoRT7W8gHdSJVl0WN1nPtRJBbWicr5nkltJqArdqNGnM0fPSIEqVT5ifQvBq1gCjAfFksysxz/jgYfShKFUf7c+U41erH9SFZkx4e9JCKTUn+iVAnWqD82bRD2smiWQuqLEZ0kFXvxLhSo89O+W1RG3TWA8cC4i/hkPPPQf97UuXzh79/E35yI8vWoe+VIX2q3jb1V3jr9NLwDnejty+YLa7zr+5ty2plfNjAeOxdtH44F3Elz6zLSEVMdZfcXesdNnpvUEVMdZfUWmtc5MO68ereIs3yLLivGg2KJ36vPTeOCdBHf1C33M8bc7O2HxCiO6+vl5jr8VD9RW0zV9/vEFjr/ZRIpWZDwoGuXdFflrPPBIgjM0eCmmU+MefxtR8v2/Tl9YW9US18jdEeOvewrc0KVzsXscf/tXvb+1SmVla9QST92zJmZmCAxe1rlTPbmPvxlJnX99q4LWhfbBbWrrHMxYmT/HF8gzHhi/VfQH9enxILj1XXWmuJp+fMvMOfnGA2t5Q6n461pT1aK4ywdbbyS4VExvN7+jpEJauWxuZkQkzVRgw5u60r49xzwmjS9g7aiH1PxqXHompGU5T4JYpA3vfqT2LbPGr4Y5OQRuKP72a3o1uUTPrPx21klRRvdHOvatJnWZXym2+6q27fopdzXJoZhz0j3HA0ODp4+pY+PR9E0g3px5TB8kvs5ZDRMzBSYyHkhKxfTWnoM6nbm6G/+2roO7/ZKsG1i46XXNjIRrTKtd9p9AOGIOZDXzG/NK+3ZzycGY+U3WvNJPcJ3tQMQMB+52lWrMcORaDqzfmO1b1poHY0M55pV2kutczVFzILLPDGTsswrsMyMDozmxRhOtZt095udcaQomus+2kPFg1BxKtJvhLW+YsaHc7lNAmHMT7nM1TXPC48FXZuy1/eaRn/zQXLLkoBlz2WCbaevyO5nMV+2BmMwDbnxr4PE2VdTqwBVT0BY7juWqqH1FV8xXCqr4gYaQqm/dQaagFafVwhMdD6xDFt9TcOMn2vLaWwo+7I0vqkoaygmNB9YnvCNqX/wd/XDxT/R3JW3wxDZO5CfmxFIIOCRwQxd/8Wut3lE7dss0h7Yyvaq1Dlm8pdErb2nFv+zSd9/hRuxFib/RpfafB7RzvXULOm+8vNJOb2jSSgQKEripZPSI/vmpJu2uyXVsuaDKWDhDoDywUi+sC6nn93/gSSMZNvfz58jFD9TcvF4LZ5TpgVCTenqaFFrq7hNNSHD3E2nWQWDSAoZSnT/TkaEG7apdICUv6nQXZ1JOmtVegXFVl2JD2vjUYrn8WIy91a4tz6zZo+700zBMfRM7qCVLDirWuUc1Lsb1TYIbif8P/fuNP1ZPU0hLuVSgiP8k1rPNtmjje79SU+g5LhUoiuxNJc//SDtrX9Du9ZWaUVamGcH96hX3qpw8r3Vz601jz+Kb0agT1bv0/dXzJ18tNXhSwFtPE/Ak8VijvXAHcS/y4upc1LB1xhZXZ1ytWjNtffMJzjkyakYAAQQQ8KIACc6LUaPNCCCAAAJ5BUhweYlYAAEEEEDAiwIkOC9GjTYjgAACCOQVIMHlJWIBBBBAAAEvCpDgvBg12owAAgggkFeABJeXiAUQQAABBLwoQILzYtRoMwIIIIBAXgESXF4iFkAAAQQQ8KJA1p1MvNgJ2owAAggggIAlYJrW40PHXlm3ybTPvL0QvycvkHkLmcnXSA2WAK7O7QfYOmOLqzOut8cDe+18RWnXoIwAAggg4BsBEpxvQklHEEAAAQTsAiQ4uwZlBBBAAAHfCJDgfBNKOoIAAgggYBcgwdk1KCOAAAII+EaABOebUNIRBBBAAAG7AAnOrkEZAQQQQMA3AiQ434SSjiCAAAII2AVIcHYNyggggAACvhEgwfkmlHQEAQQQQMAuQIKza1BGAAEEEPCNAAnON6GkIwgggAACdgESnF2DMgIIIICAbwRIcL4JJR1BAAEEELALuD/BjcTVUlWWfiyK9ZiJ3D9B1bd1yrD3jHJegZF4i6rGNb1tvUltXcN562IBBBBAwG0Crk9wRm+HTi0KqzWS0JA5qoHIPgVUo3DkWvrBduZon9obG7R5VaVc3xlXRX9YvR0XtSjcqkhiQKZ5TZFwjRTYp8jAaNp2tL9djWuf1aqqh1zVchqDAAIITEQg64GnE1lp6pYZVvfZAW0/3KwNCx6UdF3REyeV1KOqXjh7rBnlszR3eZ2WMwgXFhajT2f7Nunw/nVaYL0zGIzqxKG49MQOLZw99lahfM48LX/+cVXxzqEwW5ZGAAFXCJSZtkd4u/5Js0an2hpqtU0vK3G8UY95aOB1u63R1aaG6pek1qiONy71zKdht7u64r/8PhuB7X3C5VkN1zxAk5idaeuhFCEZ3Z/q6MmkAisq9YinWj6JiE3JqsPqPvOhTiqoFZXzPZPcpoSGjSCAgGcFPJQmDKX6u/WZavRi/ZOq8Cy5GxueUn+iVwrUqT40z40NpE0IIIBAwQIeSnBf6/KFs3cffyu4u6yQU2DkS11ot46/Vd05/pZzOSYigAACHhLwToIz+tVxKiHVcVZfsfev9JmqPQHVbX6aE0qKjUt9CCBQMgHvJLirX+hjjr85sKOM6Orn5zn+5oAsVSKAQGkFPJLgDA1eiunUeMffjKTOv75VQeui5eA2tXUOllbVU1u/oUvnYuMefzOSn+j1rU+kL7APbn1XnSkup/dUeGksAtNYwBsJLhXT283vKKmQVi6bmxUuo/sjHftWk7rMrxTbfVXbdv1UXYzDWU7ZEwyl4ofU/GpceiakZRWZu8Owuk/+H31r3ycyh85p92+atet/dXHHmGxIpiCAgAsFXH4d3HVF9z6rv7AGYNsrEI6o80BtzjMp09dzre7W3s6XVZs1YNsqmeJi5vUZU7z57M0NRrV36V/o1aR9lnWHmA91oHa+feKt8rC62rZodeKvx7XPsZLjk1zn6niPp24D2DpjjaszrlatmbYuv5PJfNUeiMk8UBjIAw0hVd+6G0dha06jpStqdeCKqcJo56hh5RLduofMNMKiqwgg4EWBzO+kvNgHW5tv6OIvfq3VO2rHbj9lm0NxkgKpX+sXH/877ahbxIXgk6RkdQQQmBoBl39FWQjCTSWjbTqi57SrdoHrBuHMj86F9Kzkyxq/VfSNX0ovNKo2YN0T1D0vT7u6hzFnS7DNyTLpibhOmnDcCjJtffIJzlCq82c6MtSQTm5KXtTpLs6kHHcvKGjGoDoP/VJDm6zkJiXj/1ddnElZkCALI4BAaQR8kOBuKnn+R9pZ+4J2r6/UjLIyzQjuV6/c9UmjNOGd5FbTl1/8F9V+9z9r/cI/UVnZnyj4gy+kh32w20yShtURQMD9Aj76itLd2Jkfnd3dWu+0DlfnYoWtM7a4OuNq1Zppy1tx56ypGQEEEECghAIkuBLis2kEEEAAAecESHDO2VIzAggggEAJBUhwJcRn0wgggAACzgmQ4JyzpWYEEEAAgRIKkOBKiM+mEUAAAQScEyDBOWdLzQgggAACJRQgwZUQn00jgAACCDgnQIJzzpaaEUAAAQRKKJB1J5MStoVNI4AAAgggMCkB0zTvrJ/1PDj7zDtLUZi0QOYtZCZdIRWkBXB1bkfA1hlbXJ1xtWq1bO0vvqK0a1BGAAEEEPCNAAnON6GkIwgggAACdgESnF2DMgIIIICAbwRIcL4JJR1BAAEEELALkODsGpQRQAABBHwjQILzTSjpCAIIIICAXYAEZ9egjAACCCDgGwESnG9CSUcQQAABBOwCJDi7BmUEEEAAAd8IkOB8E0o6ggACCCBgFyDB2TUoI4AAAgj4RoAE55tQ0hEEEEAAAbsACc6uQRkBBBBAwDcCJDjfhJKOIIAAAgjYBUhwdg3KCCCAAAK+EXB/ghuJq6WqLP2cH+tZP7l/gqpv65Thm7BMTUdG4i2qGtf0tvUmtXUNT02D2AoCCCBQRAHXJzijt0OnFoXVGkloyBzVQGSfAqpROHJN1sNZzdE+tTc2aPOqSrm+M0UM3OSrGlZvx0UtCrcqkhiQaV5TJFwjBfYpMjCath3tb1fj2me1quqhyW+OGhBAAIEpFsh6ovcUbz/P5obVfXZA2w83a8OCByVdV/TESSX1qKoXzh5bt3yW5i6v03IG4TyWGbONPp3t26TD+9dpgfXOYDCqE4fi0hM7tHD22FuF8jnztPz5x1XFO4cMPP5EAAEvCJSZ1segWy/XP0rd6FRbQ6226WUljjfqMQ8NvG63Nbra1FD9ktQa1fHGpZ75NOx219v/W178ja0zUcPVGVer1kxbD6UIyej+VEdPJhVYUalHPNVy5wJanJqH1X3mQ51UUCsq53smuRWn79SCAAJ+FfBQmjCU6u/WZ6rRi/VPqsKvESlJv1LqT/RKgTrVh+aVpAVsFAEEECi2gIcS3Ne6fOHs3cffiq0xXesb+VIX2q3jb1V3jr9NVwr6jQAC/hHwToIz+tVxKiHVcVZfsXe/9JmqPQHVbX6aE0qKjUt9CCBQMgHvJLirX+jjcY+/3VQy+t+0xrqmK7hNbZ2DJQP13oZHdPXz8+MffzN+q+gP6tMHb4Nb31VniqsNvRdjWozA9BTwSIIzNHgpplPjHX8b/FT/s6NBvzT/qP43/41e/+CyRqZnPO+j1zd06VxsnONvhgZPH1PHxqPp6w3fnHlMHyS+vo9tsAoCCCAw9QLeSHCpmN5ufkdJhbRy2dxspYrV2vM3f6bZqV6dOydtXbNELr/AL7sPJZliKBU/pOZX49IzIS2ryNwdylVRu1N/U1OhVHdc50bWaE31wyVpKRtFAAEEChXIHNEKXd/h5a8rujeksjkr1XQ6KenH2hh8QMG9UWV9CZl8X1vnLNP3frdM/yHIIJw3MINR7Q3O0JzQbp22Fn5vo4JlIe2NXs9YdUTJ93doTvV/1e9W1Cj4sMt3mYzW8ycCCExfAW9d6J0vTkZS59/4W63/8M912mUXgmdegJivK26bbyQ/0Rt/u0sf/vlhV10I7nVXt8XZ3h5s7RrFK+NaPMvMmjJt/fV2vDygP3thnZ7p+b2GOBciM/aT+rs8sFIvrAup5/d/4KbWk5JkZQQQmCoBfyU43VTy0mf63can9G0OwhV3HzKu6lJsSBufWszxzeLKUhsCCDgk4IsEZ91HsT792JdK/dWJR/XK91dxp5Oi7DDD6mrbNPaIohmNOlG9S99fPb8oNVMJAggg4LSAv47BOa01ifozvxueRFWsahPA1YZR5CK2RQa9VR2uzrhatWba+uITnHNc1IwAAggg4FUBEpxXI0e7EUAAAQTuKUCCuycPMxFAAAEEvCpAgvNq5Gg3AggggMA9BUhw9+RhJgIIIICAVwVIcF6NHO1GAAEEELinAAnunjzMRAABBBDwqgAJzquRo90IIIAAAvcUIMHdk4eZCCCAAAJeFci6k4lXO0K7EUAAAQQQME3zDkLWLYntM+8sRWHSApm3kJl0hVSQFsDVuR0BW2dscXXG1arVsrW/+IrSrkEZAQQQQMA3AiQ434SSjiCAAAII2AVIcHYNyggggAACvhEgwfkmlHQEAQQQQMAuQIKza1BGAAEEEPCNAAnON6GkIwgggAACdgESnF2DMgIIIICAbwRIcL4JJR1BAAEEELALkODsGpQRQAABBHwjQILzTSjpCAIIIICAXYAEZ9egjAACCCDgGwESnG9CSUcQQAABBOwCJDi7BmUEEEAAAd8IkOB8E0o6ggACCCBgFyDB2TUoI4AAAgj4RsD9CW4krpaqsvRzfqxn/eT+Caq+rVOGb8IyNR0ZibeoalzT29ab1NY1PDUNYisIIIBAEQVcn+CM3g6dWhRWayShIXNUA5F9CqhG4cg1WQ9nNUf71N7YoM2rKuX6zhQxcJOvali9HRe1KNyqSGJApnlNkXCNFNinyMBo2na0v12Na5/VqqqHJr85akAAAQSmWCDrid5TvP08mxtW99kBbT/crA0LHpR0XdETJ5XUo6peOHts3fJZmru8TssZhPNYZsw2+nS2b5MO71+nBdY7g8GoThyKS0/s0MLZY28VyufM0/LnH1cV7xwy8PgTAQS8IFBmWh+Dbr1c/yh1o1NtDbXappeVON6oxzw08Lrd1uhqU0P1S1JrVMcbl3rm07DbXW//b3nxN7bORA1XZ1ytWjNtPZQiJKP7Ux09mVRgRaUe8VTLnQtocWoeVveZD3VSQa2onO+Z5FacvlMLAgj4VcBDacJQqr9bn6lGL9Y/qQq/RqQk/UqpP9ErBepUH5pXkhawUQQQQKDYAh5KcF/r8oWzdx9/K7bGdK1v5EtdaLeOv1XdOf42XSnoNwII+EfAOwnO6FfHqYRUx1l9xd790meq9gRUt/lpTigpNi71IYBAyQS8k+CufqGPOf7mwI4yoqufn+f4mwOyVIkAAqUV8EiCMzR4KaZTeY+/3VC85T+qqiWukdK6emjrN3TpXCz/8bfUebWsqVdLPOWhvtFUBBCYzgLeSHCpmN5ufkdJhbRy2dxx4mUoFf9H7Wn61TjzmZwtYJkdUvOrcemZkJZVjLc73FD8rf+uptM3s6tgCgIIIOBSgfFGNJc097qie0Mqm7NSTaeTkn6sjcEHFNwb1WBmC1Mx/UP7PO3cvzpzDn/nEhiMam9whuaEduu0Nf+9jQqWhbQ3ej1jaSsJHlH7or/S/iUZs/gTAQQQcLGAyxPcfNUeiI3dksu6LdetnysHajMuE7ih+D9EtXjnc/q3Lr83i2v2hYpaHbjy/03HbGM6UDv/7iam3zgEtPM/VQvau2n4CwEE3C3g8gQ3MTyj63/r54u/o/Xp23lNbB2WmojAsLraP9HinQ1jt/OayCosgwACCLhEwFu36sqJllK85TmFmtJftN1ZYsnBmDr31LjmU0fmLWTuNNTNBetJDktDauqxN3K1DsZ+qT01t+4Fap9VgrInXUvgdD+bxPZ+1PKvg2t+o/tdItPWBwnOTjGW7DarxVXJzWphJry91Z4op5PdHumoe5KbL1xdHHzP77MutcXVucBk2vriK0rnuKgZAQQQQMCrAj77BOfeMGS+s3BvS73VMlydixe2ztji6oyrVWumLZ/gnLOmZgQQQACBEgqQ4EqIz6YRQAABBJwTIME5Z0vNCCCAAAIlFCDBlRCfTSOAAAIIOCdAgnPOlpoRQAABBEooQIIrIT6bRgABBBBwToAE55wtNSOAAAIIlFCABFdCfDaNAAIIIOCcAAnOOVtqRgABBBAooUDWnUxK2BY2jQACCCCAwKQErEd/3X7d9Ygv+4zbC/AbAQQQQAABLwrwFaUXo0abEUAAAQTyCpDg8hKxAAIIIICAFwVIcF6MGm1GAAEEEMgr8P8ANucLjMc9JAAAAAAASUVORK5CYII="></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><em>Show that T</em><sub>1</sub>, <em>T</em><sub>2</sub>, <em>T</em><sub>3</sub>, <em>T</em><sub>4 </sub>under the operation of composition of transformations form a group. Associativity may be assumed.</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that this group is cyclic.</p>
<div class="marks">[1]</div>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Write down the 2 × 2 matrices representing <em>T</em><sub>3</sub>, <em>T</em><sub>4</sub> and <em>T</em><sub>5</sub>.</p>
<div class="marks">[3]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the 2 × 2 matrix representing <em>T</em>.</p>
<div class="marks">[2]</div>
<div class="question_part_label">d.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Give a geometric description of the transformation <em>T</em>.</p>
<div class="marks">[1]</div>
<div class="question_part_label">d.ii.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><img src="data:image/png;base64,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"> <em><strong>A2</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<p><strong>Note:</strong> Award <em><strong>A1</strong> </em>for 6, 7 or 8 correct.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>the table is closed – no new elements <em><strong>A1</strong></em></p>
<p><em>T</em><sub>1</sub> is the identity <em><strong>A1</strong></em></p>
<p><em>T</em><sub>3</sub> (and <em>T</em><sub>1</sub>) are self-inverse; <em>T</em><sub>2</sub> and <em>T</em><sub>4</sub> are an inverse pair. Hence every element has an inverse <em><strong>A1</strong></em></p>
<p>hence it is a group <em><strong>AG</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>all elements in the group can be generated by <em>T</em><sub>2</sub> (or <em>T</em><sub>4</sub>) <em><strong>R1</strong></em></p>
<p>hence the group is cyclic <em><strong> AG</strong></em></p>
<p><em><strong>[1 mark]</strong></em></p>
<p> </p>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p> </p>
<p><em>T</em><sub>3</sub> is represented by \(\left( {\begin{array}{*{20}{c}}<br> { - 1}&0 \\ <br> 0&{ - 1} <br>\end{array}} \right)\) <em><strong>A1</strong></em></p>
<p><em>T</em><sub>4</sub> is represented by \(\left( {\begin{array}{*{20}{c}}<br> 0&{ - 1} \\ <br> 1&0 <br>\end{array}} \right)\) <em><strong>A1</strong></em></p>
<p><em>T</em><sub>5</sub> is represented by \(\left( {\begin{array}{*{20}{c}}<br> 1&0 \\ <br> 0&{ - 1} <br>\end{array}} \right)\) <em><strong>A1</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\left( {\begin{array}{*{20}{c}}<br> 0&{ - 1} \\ <br> 1&0 <br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br> 1&0 \\ <br> 0&{ - 1} <br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br> { - 1}&0 \\ <br> 0&{ - 1} <br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br> 0&{ - 1} \\ <br> { - 1}&0 <br>\end{array}} \right)\) <em><strong>(M1)A1</strong></em></p>
<p><strong>Note:</strong> Award <em><strong>M1A0</strong> </em>for multiplying the matrices in the wrong order.</p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">d.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>a reflection in the line \(y = - x\) <em><strong>A1</strong></em></p>
<p><em><strong>[1 mark]</strong></em></p>
<div class="question_part_label">d.ii.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.ii.</div>
</div>
<br><hr><br>