File "markSceme-SL-paper2.html"
Path: /IB QUESTIONBANKS/4 Fourth Edition - PAPER/HTML/Further Mathematics/Topic 1/markSceme-SL-paper2html
File size: 154.88 KB
MIME-type: text/x-tex
Charset: utf-8
<!DOCTYPE html>
<html>
<meta http-equiv="content-type" content="text/html;charset=utf-8">
<head>
<meta charset="utf-8">
<title>IB Questionbank</title>
<link rel="stylesheet" media="all" href="css/application-212ef6a30de2a281f3295db168f85ac1c6eb97815f52f785535f1adfaee1ef4f.css">
<link rel="stylesheet" media="print" href="css/print-6da094505524acaa25ea39a4dd5d6130a436fc43336c0bb89199951b860e98e9.css">
<script src="js/application-13d27c3a5846e837c0ce48b604dc257852658574909702fa21f9891f7bb643ed.js"></script>
<script type="text/javascript" async src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.5/MathJax.js?config=TeX-MML-AM_CHTML-full"></script>
<!--[if lt IE 9]>
<script src='https://cdnjs.cloudflare.com/ajax/libs/html5shiv/3.7.3/html5shiv.min.js'></script>
<![endif]-->
<meta name="csrf-param" content="authenticity_token">
<meta name="csrf-token" content="iHF+M3VlRFlNEehLVICYgYgqiF8jIFlzjGNjIwqOK9cFH3ZNdavBJrv/YQpz8vcspoICfQcFHW8kSsHnJsBwfg==">
<link href="favicon.ico" rel="shortcut icon">
</head>
<body class="teacher questions-show">
<div class="navbar navbar-fixed-top">
<div class="navbar-inner">
<div class="container">
<div class="brand">
<div class="inner"><a href="http://ibo.org/">ibo.org</a></div>
</div>
<ul class="nav">
<li>
<a href="../../index.html">Home</a>
</li>
<li class="active dropdown">
<a class="dropdown-toggle" data-toggle="dropdown" href="#">
Questionbanks
<b class="caret"></b>
</a><ul class="dropdown-menu">
<li>
<a href="../../geography.html" target="_blank">DP Geography</a>
</li>
<li>
<a href="../../physics.html" target="_blank">DP Physics</a>
</li>
<li>
<a href="../../chemistry.html" target="_blank">DP Chemistry</a>
</li>
<li>
<a href="../../biology.html" target="_blank">DP Biology</a>
</li>
<li>
<a href="../../furtherMath.html" target="_blank">DP Further Mathematics HL</a>
</li>
<li>
<a href="../../mathHL.html" target="_blank">DP Mathematics HL</a>
</li>
<li>
<a href="../../mathSL.html" target="_blank">DP Mathematics SL</a>
</li>
<li>
<a href="../../mathStudies.html" target="_blank">DP Mathematical Studies</a>
</li>
</ul></li>
<!-- - if current_user.is_language_services? && !current_user.is_publishing? -->
<!-- %li= link_to "Language services", tolk_path -->
</ul>
<ul class="nav pull-right">
<li>
<a href="https://06082010.xyz">IB Documents (2) Team</a>
</li></ul>
</div>
</div>
</div>
<div class="page-content container">
<div class="row">
<div class="span24">
</div>
</div>
<div class="page-header">
<div class="row">
<div class="span16">
<p class="back-to-list">
</p>
</div>
<div class="span8" style="margin: 0 0 -19px 0;">
<img style="width: 100%;" class="qb_logo" src="images/logo.jpg" alt="Ib qb 46 logo">
</div>
</div>
</div><h2>SL Paper 2</h2><div class="specification">
<p>Consider the matrix</p>
<p style="text-align: center;"><em><strong>A</strong></em> \( = \left[ {\begin{array}{*{20}{c}} \lambda &3&2 \\ 2&4&\lambda \\ 3&7&3 \end{array}} \right]\).</p>
</div>
<div class="specification">
<p>Suppose now that \(\lambda = 1\) so consider the matrix</p>
<p style="text-align: center;"><strong><em>B</em></strong> \( = \left[ {\begin{array}{*{20}{c}} 1&3&2 \\ 2&4&1 \\ 3&7&3 \end{array}} \right]\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find an expression for det(<strong><em>A</em></strong>) in terms of \(\lambda \), simplifying your answer.</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>Hence show that <strong><em>A </em></strong>is singular when \(\lambda = 1\) and find the other value of \(\lambda \) for which <strong><em>A </em></strong>is singular.</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>Explain how it can be seen immediately that <strong><em>B </em></strong>is singular without calculating its determinant.</p>
<div class="marks">[1]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Determine the null space of <strong><em>B</em></strong>.</p>
<div class="marks">[4]</div>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Explain briefly how your results verify the rank-nullity theorem.</p>
<div class="marks">[[N/A]]</div>
<div class="question_part_label">b.iii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Prove, using mathematical induction, that</p>
<p style="text-align: center;"><strong><em>B</em></strong>\(^n = {8^{n - 2}}\)<strong><em>B</em></strong>\(^2\) for \(n \in {\mathbb{Z}^ + },{\text{ }}n \geqslant 3\).</p>
<div class="marks">[7]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>det(<strong><em>A</em></strong>) \( = \lambda (12 - 7\lambda ) + 3(3\lambda - 6) + 2(14 - 12)\) <strong><em>M1A1</em></strong></p>
<p>\( = 12\lambda - 7{\lambda ^2} + 9\lambda - 18 + 4\)</p>
<p>\( = - 7{\lambda ^2} + 21\lambda - 14\) <strong><em>A1</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong><em>A </em></strong>is singular when \(\lambda = 1\) because the determinant is zero <strong><em>R1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> Do not award the <strong><em>R1 </em></strong>if the determinant has not been obtained.</p>
<p> </p>
<p>the other value is 2 <strong><em>A1</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>the third row is the sum of the first two rows <strong><em>A1</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>the null space satisfies</p>
<p>\(\left[ {\begin{array}{*{20}{c}} 1&3&2 \\ 2&4&1 \\ 3&7&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0 \\ 0 \\ 0 \end{array}} \right]\) <strong><em>M1</em></strong></p>
<p>\(x + 3y + 2z = 0\)</p>
<p>\(2x + 4y + z = 0\) <strong><em>(A1)</em></strong></p>
<p>\(3x + 7y + 3z = 0\)</p>
<p>the solution is (by GDC or otherwise)</p>
<p>\(\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5 \\ { - 3} \\ 2 \end{array}} \right]\alpha \) where \(\alpha \in \mathbb{R}\) <strong><em>M1A1</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>the rank-nullity theorem for square matrices states that</p>
<p>rank of matrix + dimension of null space = number of columns <strong><em>A1</em></strong></p>
<p>here, rank = 2, dimension of null space = 1 and number of columns = 3 <strong><em>A1</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.iii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>first show that the result is true for \(n = 3\)</p>
<p><strong><em>B</em></strong>\(^2 = \left[ {\begin{array}{*{20}{c}} {13}&{29}&{11} \\ {13}&{29}&{11} \\ {26}&{58}&{22} \end{array}} \right]\) <strong><em>A1</em></strong></p>
<p><strong><em>B</em></strong>\(^3 = \left[ {\begin{array}{*{20}{c}} {104}&{232}&{88} \\ {104}&{232}&{88} \\ {208}&{464}&{176} \end{array}} \right]\) <strong><em>A1</em></strong></p>
<p>therefore <strong><em>B</em></strong>\(^3 = 8\)<strong><em>B</em></strong>\(^2\) so true for \(n = 3\) <strong><em>R1</em></strong></p>
<p>assume the result is true for \(n = k\), that is <strong><em>B</em></strong>\(^k = {8^{k - 2}}\)<strong><em>B</em></strong>\(^2\) <strong><em>M1</em></strong></p>
<p>consider <strong><em>B</em></strong>\(^{k + 1} = {8^{k - 2}}\)<strong><em>B</em></strong>\(^2\) <strong><em>M1</em></strong></p>
<p>\( = {8^{k - 2}}8\)<strong><em>B</em></strong>\(^2\)</p>
<p>\( = {8^{k - 1}}\)<strong><em>B</em></strong>\(^2\) <strong><em>A1</em></strong></p>
<p>therefore, true for \(n = k \Rightarrow \) true for \(n = k + 1\) and since the result is true for \(n = 3\), it is true for \(n \geqslant 3\) <strong><em>R1</em></strong></p>
<p><strong><em>[7 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.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>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.iii.</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 hyperbola with equation \({x^2} - 4xy - 2{y^2} = 3\) is rotated through an acute anticlockwise angle \(\alpha \) about the origin.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>The point \((x,{\text{ }}y)\) is rotated through an anticlockwise angle \(\alpha \) about the origin to become the point \((X,{\text{ }}Y)\). Assume that the rotation can be represented by</p>
<p>\[\left[ {\begin{array}{*{20}{c}} X \\ Y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right].\]</p>
<p>Show, by considering the images of the points \((1,{\text{ }}0)\) and \((0,{\text{ }}1)\) under this rotation that</p>
<p>\[\left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} {\cos \alpha }&{ - \sin \alpha } \\ {\sin \alpha }&{\cos \alpha } \end{array}} \right].\]</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>By expressing \((x,{\text{ }}y)\) in terms of \((X,{\text{ }}Y)\), determine the equation of the rotated hyperbola in terms of \(X\) and \(Y\).</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>Verify that the coefficient of \(XY\) in the equation is zero when \(\tan \alpha = \frac{1}{2}\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Determine the equation of the rotated hyperbola in this case, giving your answer in the form \(\frac{{{X^2}}}{{{A^2}}} - \frac{{{Y^2}}}{{{B^2}}} = 1\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.iii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Hence find the coordinates of the foci of the hyperbola prior to rotation.</p>
<div class="marks">[5]</div>
<div class="question_part_label">b.iv.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>consider \(\left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1 \\ 0 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} a \\ c \end{array}} \right]\) <strong><em>(M1)</em></strong></p>
<p>the image of \((1,{\text{ }}0)\) is \((\cos \alpha ,{\text{ }}\sin \alpha )\) <strong><em>A1</em></strong></p>
<p>therefore \(a = \cos \alpha ,{\text{ }}c = \sin \alpha \) <strong><em>AG</em></strong></p>
<p>consider \(\left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0 \\ 1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} b \\ d \end{array}} \right]\)</p>
<p>the image of \((0,{\text{ }}1)\) is \(( - \sin \alpha ,{\text{ }}\cos \alpha )\) <strong><em>A1</em></strong></p>
<p>therefore \(b = - \sin \alpha ,{\text{ }}d = \cos \alpha \) <strong><em>AG</em></strong></p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\left[ {\begin{array}{*{20}{c}} X \\ Y \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} {\cos \alpha }&{ - \sin \alpha } \\ {\sin \alpha }&{\cos \alpha } \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] \Rightarrow \left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha } \\ { - \sin \alpha }&{\cos \alpha } \end{array}} \right]\left[ {\begin{array}{*{20}{c}} X \\ Y \end{array}} \right]\)</p>
<p>or \(x = X\cos \alpha + Y\sin \alpha ,{\text{ }}y = - X\sin \alpha + Y\cos \alpha \) <strong><em>A1</em></strong></p>
<p>substituting in the equation of the hyperbola, <strong><em>M1</em></strong></p>
<p>\({(X\cos \alpha + Y\sin \alpha )^2} - 4(X\cos \alpha + Y\sin \alpha )( - X\sin \alpha + Y\cos \alpha )\)</p>
<p>\( - 2{( - X\sin \alpha + Y\cos \alpha )^2} = 3\) <strong><em>A1</em></strong></p>
<p>\({X^2}({\cos ^2}\alpha - 2{\sin ^2}\alpha + 4\sin \alpha \cos \alpha ) + \)</p>
<p>\(XY(2\sin \alpha \cos \alpha - 4{\cos ^2}\alpha + 4{\sin ^2}\alpha + 4\sin \alpha \cos \alpha ) + \)</p>
<p>\({Y^2}({\sin ^2}\alpha - 2{\cos ^2}\alpha - 4\sin \alpha \cos \alpha ) = 3\)</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>when \(\tan \alpha = \frac{1}{2},{\text{ }}\sin \alpha = \frac{1}{{\sqrt 5 }}\) and \(\cos \alpha = \frac{2}{{\sqrt 5 }}\) <strong><em>A1</em></strong></p>
<p>the \(XY{\text{ term}} = 6\sin \alpha \cos \alpha - 4{\cos ^2}\alpha + 4{\sin ^2}\alpha \) <strong><em>M1</em></strong></p>
<p>\( = 6 \times \frac{1}{{\sqrt 5 }} \times \frac{2}{{\sqrt 5 }} - 4 \times \frac{4}{5} + 4 \times \frac{1}{5}\left( {\frac{{12}}{5} - \frac{{16}}{5} + \frac{4}{5}} \right)\) <strong><em>A1</em></strong></p>
<p>\( = 0\) <strong><em>AG</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>the equation of the rotated hyperbola is</p>
<p>\(2{X^2} - 3{Y^2} = 3\) <strong><em>M1A1</em></strong></p>
<p>\(\frac{{{X^2}}}{{{{\left( {\sqrt {\frac{3}{2}} } \right)}^2}}} - \frac{{{Y^2}}}{{{{(1)}^2}}} = 1\) <strong><em>A1</em></strong></p>
<p>\(\left( {{\text{accept }}\frac{{{X^2}}}{{\frac{3}{2}}} - \frac{{{Y^2}}}{1} = 1} \right)\)</p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.iii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>the coordinates of the foci of the rotated hyperbola</p>
<p>are \(\left( { \pm \sqrt {\frac{3}{2} + 1} ,{\text{ }}0} \right) = \left( { \pm \sqrt {\frac{5}{2}} ,{\text{ }}0} \right)\) <strong><em>M1A1</em></strong></p>
<p>the coordinates of the foci prior to rotation were given by</p>
<p>\(\left[ {\begin{array}{*{20}{c}} {\frac{2}{{\sqrt 5 }}}&{\frac{1}{{\sqrt 5 }}} \\ { - \frac{1}{{\sqrt 5 }}}&{\frac{2}{{\sqrt 5 }}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { \pm \sqrt {\frac{5}{2}} } \\ 0 \end{array}} \right]\)</p>
<p><strong><em>M1A1</em></strong></p>
<p>\(\left[ {\begin{array}{*{20}{c}} { \pm \sqrt 2 } \\ { \mp \frac{1}{{\sqrt 2 }}} \end{array}} \right]\) <strong><em>A1</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.iv.</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">b.iii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.iv.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">\(S\) is defined as the set of all \(2 \times 2\) <span class="s1">non-singular matrices. </span>\(A\) <span class="s1">and </span>\(B\) <span class="s1">are two elements of the set </span>\(S\)<span class="s1">.</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Show that \({({A^T})^{ - 1}} = {({A^{ - 1}})^T}\).</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Show that \({(AB)^T} = {B^T}{A^T}\).</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">A relation \(R\) is defined on \(S\) such that \(A\) is related to \(B\) if and only if there exists an element \(X\) of \(S\) such that \(XA{X^T} = B\). Show that \(R\) is an equivalence relation.</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 class="p1">(i) <span class="Apple-converted-space"> </span>\(A = \left( {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right)\)</p>
<p class="p1">\({A^T} = \left( {\begin{array}{*{20}{c}} a&c \\ b&d \end{array}} \right)\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\({({A^T})^{ - 1}} = \frac{1}{{ad - bc}}\left( {\begin{array}{*{20}{c}} d&{ - c} \\ { - b}&a \end{array}} \right)\;\;\;\)(which exists because \(ad - bc \ne 0\)) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\({A^{ - 1}} = \frac{1}{{ad - bc}}\left( {\begin{array}{*{20}{c}} d&{ - b} \\ { - c}&a \end{array}} \right)\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\({({A^{ - 1}})^T} = \frac{1}{{ad - bc}}\left( {\begin{array}{*{20}{c}} d&{ - c} \\ { - b}&a \end{array}} \right)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">hence \({({A^T})^{ - 1}} = {({A^{ - 1}})^T}\) as required <strong><em>AG</em></strong></p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>\(A = \left( {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right)\;\;\;B = \left( {\begin{array}{*{20}{c}} e&f \\ g&h \end{array}} \right)\)</p>
<p class="p1">\(AB = \left( {\begin{array}{*{20}{c}} {ae + bg}&{af + bh} \\ {ce + dg}&{cf + dh} \end{array}} \right)\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\({(AB)^T} = \left( {\begin{array}{*{20}{c}} {ae + bg}&{ce + dg} \\ {af + bh}&{cf + dh} \end{array}} \right)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\({B^T} = \left( {\begin{array}{*{20}{c}} e&g \\ f&h \end{array}} \right)\;\;\;{A^T} = \left( {\begin{array}{*{20}{c}} a&c \\ b&d \end{array}} \right)\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\({B^T}{A^T} = \left( {\begin{array}{*{20}{c}} e&g \\ f&h \end{array}} \right)\left( {\begin{array}{*{20}{c}} a&c \\ b&d \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {ae + bg}&{ce + dg} \\ {af + bh}&{cf + dh} \end{array}} \right)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">hence \({(AB)^T} = {B^T}{A^T}\) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">\(R\) is reflexive since \(I \in S\) and \(IA{I^T} = A\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\(XA{X^T} = B \Rightarrow A = {X^{ - 1}}B{({X^T})^{ - 1}}\) <span class="Apple-converted-space"> </span><strong><em>M1A1</em></strong></p>
<p class="p1">\( \Rightarrow A = {X^{ - 1}}B{({X^{ - 1}})^T}\) from a (i) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">which is of the correct form, hence symmetric <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1">\(ARB \Rightarrow XA{X^T} = B\) and \(BRC = YB{Y^T} = C\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1"><strong>Note:<span class="Apple-converted-space"> </span></strong>Allow use of \(X\) rather than \(Y\) in this line.</p>
<p class="p2"> </p>
<p class="p1">\( \Rightarrow YXA{X^T}{Y^T} = YB{Y^T} = C\) <span class="Apple-converted-space"> </span><strong><em>M1A1</em></strong></p>
<p class="p1">\( \Rightarrow (YX)A{(YX)^T} = C\) from a (ii) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">this is of the correct form, hence transitive</p>
<p class="p1">hence \(R\) is an equivalence relation <span class="Apple-converted-space"> </span><strong><em>AG</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">Part a) was successfully answered by the majority of candidates..</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">There were some wholly correct answers seen to part b) but a number of candidates struggled with the need to formally explain what was required.</p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p><span style="font-family: 'times new roman', times; font-size: medium;">The matrix <em><strong>A</strong></em> is given by <em><strong>A</strong></em> = \(\left( {\begin{array}{*{20}{c}}1&2&3&4\\3&8&{11}&8\\1&3&4&\lambda \\\lambda &5&7&6\end{array}} \right)\).</span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">(a) Given that \(\lambda = 2\), </span><em style="font-family: 'times new roman', times; font-size: medium;"><strong>B = </strong></em><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( \begin{array}{l}2\\4\\\mu \\3\end{array} \right)\) and <em><strong>X</strong></em> = \(\left( \begin{array}{l}x\\y\\z\\t\end{array} \right)\),</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;">(i) find the value of \(\mu \) for which the equations defined by <strong><em>AX </em></strong>= <strong><em>B </em></strong>are consistent and solve the equations in this case;</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) define the rank of a matrix and state the rank o</span><span style="font-family: 'times new roman', times; font-size: medium;">f <em><strong>A</strong></em>.</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 \(\lambda = 1\),</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;">(i) show that the four column vectors in <strong><em>A </em></strong>form a basis for the space of four-dimensional column vectors;</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) express the vector \(\left( \begin{array}{c}6\\28\\12\\15\end{array} \right)\) as a linear combination of these basis vectors.</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) (i) using row reduction, <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;">\(\left( {\begin{array}{*{20}{c}}1&2&3&4\\3&8&{11}&8\\1&3&4&2\\2&5&7&6\end{array}{\rm{\,\,\,\,\,\,\,\,}}\begin{array}{*{20}{c}}2\\4\\\mu \\3\end{array}} \right)\)</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;">\(\left( {\begin{array}{*{20}{c}}1&2&3&4\\0&2&2&{ - 4}\\0&1&1&{ - 2}\\0&1&1&{ - 2}\end{array}{\rm{\,\,\,\,\,\,\,\,}}\begin{array}{*{20}{c}}2\\{ - 2}\\{\mu - 2}\\{ - 1}\end{array}} \right)\) <strong><em>(A2)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">for consistency,</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\mu - 2 = - 1\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\mu = 1\) <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;">put \(z = \alpha ,{\text{ }}t = \beta \) <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;">\(y = - 1 - \alpha + 2\beta ;{\text{ }}x = 4 - \alpha - 8\beta \) <strong><em>A1A1</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) the rank of a matrix is the number of independent rows (or columns) <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;">\({\text{rank}}\,{\text{(}}\)<strong><em>A</em></strong>\() = 2\) <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;"><strong><em>[10 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) \({\text{det}}\,{\text{(}}\)<strong><em>A</em></strong>\() = 2\) <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;">since \({\text{det}}\,{\text{(}}\)<strong><em>A</em></strong>\() \ne 0\), the vectors form a basis <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;">(ii) let \(\left( \begin{array}{l}6\\28\\12\\15\end{array} \right) = a\left( \begin{array}{l}1\\3\\1\\1\end{array} \right) + b\left( \begin{array}{l}2\\8\\3\\5\end{array} \right) + c\left( \begin{array}{c}3\\11\\4\\7\end{array} \right) + d\left( \begin{array}{l}4\\8\\1\\6\end{array} \right)\) <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;">\( = \left( {\begin{array}{*{20}{c}}1&2&3&4\\3&8&{11}&8\\1&3&4&1\\1&5&7&6\end{array}} \right)\left( \begin{array}{l}a\\b\\c\\d\end{array} \right)\)</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 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;">\(\left( \begin{array}{l}a\\b\\c\\d\end{array} \right) = {\left( {\begin{array}{*{20}{c}}1&2&3&4\\3&8&{11}&8\\1&3&4&1\\1&5&7&6\end{array}} \right)^{ - 1}}\left( \begin{array}{l}6\\28\\12\\15\end{array} \right)\)</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;">\(= \left( \begin{array}{c}2\\1\\2\\ - 1\end{array} \right)\)</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</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;">\(a = 2\) <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;">\(b = 1\) <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;">\(c = 2\) <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;">\(d = - 1\) <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;">\(\left( \begin{array}{l}6\\28\\12\\15\end{array} \right) = 2\left( \begin{array}{l}1\\3\\1\\1\end{array} \right) + \left( \begin{array}{l}2\\8\\3\\5\end{array} \right) + 2\left( \begin{array}{c}3\\11\\4\\7\end{array} \right) - \left( \begin{array}{l}4\\8\\1\\6\end{array} \right)\)</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>[8 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">The set of all permutations of the list of the integers \(1,{\text{ }}2,{\text{ }}3{\text{ }} \ldots {\text{ }}n\) is a group, \({S_n}\), under the operation of composition of permutations.</p>
</div>
<div class="specification">
<p class="p1"><span class="s1">Each element of \({S_4}\) </span>can be represented by a \(4 \times 4\) matrix. For example, the cycle \({\text{(1 2 3 4)}}\) is represented by the matrix</p>
<p class="p1">\(\left( {\begin{array}{*{20}{c}} 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \\ 1&0&0&0 \end{array}} \right)\) acting on the column vector \(\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ 3 \\ 4 \end{array}} \right)\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1"><span class="s1">(i) <span class="Apple-converted-space"> </span></span>Show that the order of \({S_n}\) is \(n!\);</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>List the <span class="s2">6 </span>elements of \({S_3}\) <span class="s1">in cycle form;</span></p>
<p class="p1">(iii) <span class="Apple-converted-space"> </span>Show that \({S_3}\) <span class="s1">is not Abelian;</span></p>
<p class="p1">(iv) <span class="Apple-converted-space"> </span>Deduce that \({S_n}\) is not Abelian for \(n \geqslant 3\).</p>
<div class="marks">[9]</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>Write down the matrices <span class="s1"><strong><em>M</em></strong>\(_1\), <strong><em>M</em></strong>\(_2\)</span> representing the permutations \((1{\text{ }}2),{\text{ }}(2{\text{ }}3)\)<span class="s1">, </span><span class="s3">respectively;</span></p>
<p class="p2"><span class="s4">(ii) <span class="Apple-converted-space"> </span>Find </span><span class="s1"><strong><em>M</em></strong>\(_1\)<strong><em>M</em></strong>\(_2\)</span> and state the permutation represented by this matrix;</p>
<p class="p1"><span class="s3">(</span>iii) <span class="Apple-converted-space"> </span>Find \(\det (\)<span class="s1"><strong><em>M</em></strong></span><span class="s5">\(_1)\)</span><span class="s1">, \(\det (\)<strong><em>M</em></strong></span><span class="s5">\(_2)\) </span>and deduce the value of \(\det (\)<span class="s1"><strong><em>M</em></strong>\(_1\)<strong><em>M</em></strong></span>\(_2)\)<span class="s5">.</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1"><span class="s1">(i) <span class="Apple-converted-space"> </span></span>Use mathematical induction to prove that</p>
<p class="p1"><span class="s1">\((1{\text{ }}n)(1{\text{ }}n{\text{ }} - 1)(1{\text{ }}n - 2) \ldots (1{\text{ }}2) = (1{\text{ }}2{\text{ }}3 \ldots n){\text{ }}n \in {\mathbb{Z}^ + },{\text{ }}n > 1\).</span></p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Deduce that every permutation can be written as a product of cycles of length 2.</p>
<div class="marks">[8]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="s1">(i) <span class="Apple-converted-space"> </span>1 </span>has \(n\) <span class="s1">possible new positions; 2 </span>then has \(n - 1\) possible new positions…</p>
<p class="p1">\(n\) has only one possible new position <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">the number of possible permutations is \(n \times (n - 1) \times \ldots \times 2 \times 1\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\( = n!\) </span><strong><em>AG</em></strong></p>
<p class="p2"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Give no credit for simply stating that the number of permutations is \(n!\)</p>
<p class="p2"> </p>
<p class="p3">(ii) <span class="Apple-converted-space"> \((1)(2)(3);{\text{ }}(1{\text{ }}2)(3);{\text{ }}(1{\text{ }}3)(2);{\text{ }}(2{\text{ }}3)(1);{\text{ }}(1{\text{ }}2{\text{ }}3);{\text{ }}(1{\text{ }}3{\text{ }}2)\)</span> <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A2</em></strong></span></p>
<p class="p2"> </p>
<p class="p1"><strong>Notes: <em>A1 </em></strong>for 4 or 5 correct.</p>
<p class="p1">If single bracket terms are missing, do not penalize.</p>
<p class="p1">Accept \(e\) in place of the identity.</p>
<p class="p2"> </p>
<p class="p1">(iii) <span class="Apple-converted-space"> </span>attempt to compare \({\pi _1} \circ {\pi _2}\) <span class="s1">with \({\pi _2} \circ {\pi _1}\) </span>for two permutations <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p3">for example \((1{\text{ }}2)(1{\text{ }}3) = (1{\text{ }}3{\text{ }}2)\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p3">but \((1{\text{ }}3)(1{\text{ }}2) = (1{\text{ }}2{\text{ }}3)\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p1">hence \({S_3}\) is not Abelian <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1"> </p>
<p class="p1">(iv) <span class="Apple-converted-space"> \({S_3}\)</span> is a subgroup of \({S_n}\), <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">so \({S_n}\) contains non-commuting elements <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">\( \Rightarrow {S_n}\) is not Abelian for \(n \geqslant 3\) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1"><strong><em>[9 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><span class="s1"><strong><em>M</em></strong>\(_1 = \left( {\begin{array}{*{20}{c}} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{array}} \right)\)</span>, <span class="s1"><strong><em>M</em></strong>\(_2 = \left( {\begin{array}{*{20}{c}} 1&0&0&0 \\ 0&0&1&0 \\ 0&1&0&0 \\ 0&0&0&1 \end{array}} \right)\) <span class="Apple-converted-space"> </span></span><strong><em>A1A1</em></strong></p>
<p class="p2"> </p>
<p class="p2"><span class="s2">(ii) <span class="Apple-converted-space"> </span></span><strong><em>M</em></strong>\(_1\)<strong><em>M</em></strong>\(_2 = \left( {\begin{array}{*{20}{c}} 0&0&1&0 \\ 1&0&0&0 \\ 0&1&0&0 \\ 0&0&0&1 \end{array}} \right)\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p2">this represents \((1{\text{ }}3{\text{ }}2)\) <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>by, for example, interchanging a pair of rows <span class="Apple-converted-space"> </span><span class="s3">(</span><strong><em>M1</em></strong><span class="s3">)</span></p>
<p class="p2">\(\det (\)<strong><em>M</em></strong>\(_1) = \det (\)<strong><em>M</em></strong>\(_2) = - 1\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p1">then \(\det (\)<span class="s1"><strong><em>M</em></strong></span>\(_1\)<span class="s1"><strong><em>M</em></strong></span>\(_2) = ( - 1) \times ( - 1) = 1\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><strong><em>[7 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>let \({\text{P}}(n)\) be the proposition that</p>
<p class="p1">\((1{\text{ }}n)(1{\text{ }}n - 1)(1{\text{ }}n - 2) \ldots (1{\text{ }}2) = (1{\text{ }}2{\text{ }}3 \ldots n){\text{ }}n \in {\mathbb{Z}^ + }\)</p>
<p class="p1">the statement that \({\text{P}}(2)\) <span class="s1">is true <em>eg</em> \((1{\text{ }}2) = (1{\text{ }}2)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></span></p>
<p class="p2"><span class="s2">assume \({\text{P}}(k)\) </span>is true for some \(k\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2">consider \((1{\text{ }}k + 1)(1{\text{ }}k)(1{\text{ }}k - 1)(1{\text{ }}k - 2) \ldots (1{\text{ }}2)\)</p>
<p class="p2"><span class="Apple-converted-space">\( = (1{\text{ }}k + 1)(1{\text{ }}2{\text{ }}3 \ldots k)\) </span><strong><em>M1</em></strong></p>
<p class="p2">then the composite permutation has the following effect on the first \(k + 1\) <span class="s2">integers: \(1 \to 2,{\text{ }}2 \to 3 \ldots k - 1 \to k,{\text{ }}k \to 1 \to k + 1,{\text{ }}k + 1 \to 1\) <span class="Apple-converted-space"> </span></span><strong><em>A1</em></strong></p>
<p class="p1">this is \((1{\text{ }}2{\text{ }}3 \ldots k{\text{ }}k + 1)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2">hence the assertion is true by induction <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p2"> </p>
<p class="p2">(ii) <span class="Apple-converted-space"> </span>every permutation is a product of cycles <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p2">generalizing the result in (i) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p2">every cycle is a product of cycles of length 2 <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p2">hence every permutation can be written as a product of cycles of length 2 <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p2"><strong><em>[8 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;">
<p class="p1">In part (a)(i), many just wrote down \(n!\) without showing how this arises by a sequential choice process. Part (ii) was usually correctly answered, although some gave their answers in the unwanted 2-dimensional form. Part (iii) was often well answered, though some candidates failed to realise that they need to explicitly evaluate the product of two elements in both orders.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Part (b) was often well answered. A number of candidates found \(2 \times 2\) matrices – this gained no marks.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Nearly all candidates knew how to approach part (c)(i), but failed to be completely convincing. Few candidates seemed to know that every permutation can be written as a product of non-overlapping cycles, as the first step in part (ii).</p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Given that the elements of a \(2 \times 2\) symmetric matrix are real, show that</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) the eigenvalues are real;</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (ii) the eigenvectors are orthogonal if the eigenvalues are distinct.</span></p>
<div class="marks">[11]</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;">The matrix \(\boldsymbol{A}\) is given by\[\boldsymbol{A} = \left( {\begin{array}{*{20}{c}}<br>{11}&{\sqrt 3 }\\<br>{\sqrt 3 }&9<br>\end{array}} \right) .\]Find the eigenvalues and eigenvectors of \(\boldsymbol{A}\).</span></p>
<div class="marks">[7]</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;">The ellipse \(E\) has equation \({{\boldsymbol{X}}^T}{\boldsymbol{AX}} = 24\) where \(\boldsymbol{X} = \left( \begin{array}{l}<br>x\\<br>y<br>\end{array} \right)\) and \(\boldsymbol{A}\) is as defined in </span><span style="font-family: times new roman,times; font-size: medium;">part (b).</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (i) Show that \(E\) can be rotated about the origin onto the ellipse \(E'\) having </span><span style="font-family: times new roman,times; font-size: medium;">equation \(2{x^2} + 3{y^2} = 6\) .</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (ii) Find the acute angle through which \(E\) has to be rotated to coincide with \(E'\) .</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">c.</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;">(i) let </span><span style="font-family: times new roman,times; font-size: medium;">\(\boldsymbol{M} = \left( {\begin{array}{*{20}{c}}<br>a&b\\<br>b&c<br>\end{array}} \right)\) <strong><em>(M1)</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the eigenvalues satisfy</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\det (\boldsymbol{M} - \lambda \boldsymbol{I}) = 0\) <strong><em>(M1)</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\((a - \lambda )(c - \lambda ) - {b^2} = 0\) <em><strong>(A1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({\lambda ^2} - \lambda (a + c) + ac - {b^2} = 0\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">discriminant \( = {(a + c)^2} - 4(ac - {b^2})\) <strong><em>M1</em></strong></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> \( = {(a - c)^2} + 4{b^2} \ge 0\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">this shows that the eigenvalues are real <em><strong>AG</strong></em></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) let the distinct eigenvalues be \({\lambda _1},{\lambda _2}\) , with eigenvectors </span><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: times new roman,times; font-size: medium;">\({{\boldsymbol{X}}_1}\)</span>, \({{\boldsymbol{X}}_2}\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">then</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({\lambda _1}{{\boldsymbol{X}}_1} = {\boldsymbol{M}}{{\boldsymbol{X}}_1}\) and </span><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: times new roman,times; font-size: medium;">\({\lambda _2}{{\boldsymbol{X}}_2} = {\boldsymbol{M}}{{\boldsymbol{X}}_1}\)</span> <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">transpose the first equation and postmultiply by \({{\boldsymbol{X}}_2}\) to give</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({\lambda _1}{\boldsymbol{X}}_1^T{{\boldsymbol{X}}_2} = {\boldsymbol{X}}_1^T{\boldsymbol{M}}{{\boldsymbol{X}}_2}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1</span></strong></em></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">premultiply the second equation by \({\boldsymbol{X}}_1^T\) <br></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: times new roman,times; font-size: medium;">\({\lambda _2}{\boldsymbol{X}}_1^T{{\boldsymbol{X}}_2} = {\boldsymbol{X}}_1^T{\boldsymbol{M}}{{\boldsymbol{X}}_2}\)</span> </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;">it follows that</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\((\lambda 1 - {\lambda _2}){\boldsymbol{X}}_1^T{{\boldsymbol{X}}_2} = 0\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">since \(\lambda 1 \ne {\lambda _2}\) </span><span style="font-family: times new roman,times; font-size: medium;">, it follows that \({\boldsymbol{X}}_1^T{{\boldsymbol{X}}_2} = 0\)</span><span style="font-family: times new roman,times; font-size: medium;"> so that the eigenvectors </span><span style="font-family: times new roman,times; font-size: medium;">are orthogonal <strong><em>R1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> </span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[11 marks]</span></em></strong></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;">the eigenvalues satisfy \(\left| \begin{array}{l}<br>11 - \lambda \\<br>\sqrt 3 <br>\end{array} \right.\left. \begin{array}{l}<br>\sqrt 3 \\<br>9 - \lambda <br>\end{array} \right| = 0\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1A1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\lambda ^2} - 20\lambda + 96 = 0\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\lambda = 8,12\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">first eigenvector satisfies</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}<br>3&{\sqrt 3 }\\<br>{\sqrt 3 }&1<br>\end{array}} \right)\left( \begin{array}{l}<br>x\\<br>y<br>\end{array} \right) = \left( \begin{array}{l}<br>0\\<br>0<br>\end{array} \right)\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1</span></em></strong></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\left( \begin{array}{l}<br>x\\<br>y<br>\end{array} \right) = \) (any multiple of) \(\left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - \sqrt 3 }<br>\end{array}} \right)\) </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,times; font-size: medium;">second eigenvector satisfies</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> \(\left( {\begin{array}{*{20}{c}}<br>{ - 1}&{\sqrt 3 }\\<br>{\sqrt 3 }&{ - 3}<br>\end{array}} \right)\left( \begin{array}{l}<br>x\\<br>y<br>\end{array} \right) = \left( \begin{array}{l}<br>0\\<br>0<br>\end{array} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left( \begin{array}{l}<br>x\\<br>y<br>\end{array} \right) = \) (any multiple of ) \(\left( {\begin{array}{*{20}{c}}<br>{\sqrt 3 }\\<br>1<br>\end{array}} \right)\) </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;"> [7 marks]</span></strong></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) consider the rotation in which \((x,y)\) is transformed onto \((x',y')\) defined by</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\left( \begin{array}{l}<br>{x'}\\<br>{y'}<br>\end{array} \right) = \left( {\begin{array}{*{20}{c}}<br>{\frac{1}{2}}&{ - \frac{{\sqrt 3 }}{2}}\\<br>{\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}<br>\end{array}} \right)\left( \begin{array}{l}<br>x\\<br>y<br>\end{array} \right)\) </span><span style="font-family: times new roman,times; font-size: medium;">so that \(\left( \begin{array}{l}<br>x\\<br>y<br>\end{array} \right) = \left( {\begin{array}{*{20}{c}}<br>{\frac{1}{2}}&{\frac{{\sqrt 3 }}{2}}\\<br>{ - \frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}<br>\end{array}} \right)\left( \begin{array}{l}<br>{x'}\\<br>{y'}<br>\end{array} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1A1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">the ellipse \(E\) becomes</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}<br>{x'}&{y'}<br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br>{\frac{1}{2}}&{ - \frac{{\sqrt 3 }}{2}}\\<br>{\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}<br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br>{11}&{\sqrt 3 }\\<br>{\sqrt 3 }&9<br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br>{\frac{1}{2}}&{\frac{{\sqrt 3 }}{2}}\\<br>{ - \frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}<br>\end{array}} \right)\left( \begin{array}{l}<br>{x'}\\<br>{y'}<br>\end{array} \right) = 24\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1A1</span></em></strong></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}<br>{x'}&{y'}<br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br>8&0\\<br>0&{12}<br>\end{array}} \right)\left( \begin{array}{l}<br>{x'}\\<br>{y'}<br>\end{array} \right) = 24\) </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,times; font-size: medium;">\(2{(x')^2} + 3{(y')^2} = 6\) <em><strong>AG</strong></em></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) the angle of rotation is given by \(\cos \theta = \frac{1}{2},\sin \theta = \frac{{\sqrt 3 }}{2}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">since a rotational matrix has the form \(\left( {\begin{array}{*{20}{c}}<br>{\cos \theta }&{ - \sin \theta }\\<br>{\sin \theta }&{\cos \theta }<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">so \(\theta = {60^ \circ }\) (anticlockwise) <strong><em>A1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> [7 marks]</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The function \(f:\mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}\)</span><span style="font-family: times new roman,times; font-size: medium;"> is defined by \(\boldsymbol{X} \mapsto \boldsymbol{AX}\) , where \(\boldsymbol{X} = \left[ \begin{array}{l}<br>x\\<br>y<br>\end{array} \right]\) and \(\boldsymbol{A} = \left[ \begin{array}{l}<br>a\\<br>c<br>\end{array} \right.\left. \begin{array}{l}<br>b\\<br>d<br>\end{array} \right]\) </span><span style="font-family: times new roman,times; font-size: medium;">where \(a\) , \(b\) , \(c\) , \(d\) are all non-zero.</span></p>
</div>
<div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Consider the group \(\left\{ {S,{ + _m}} \right\}\) where \(S = \left\{ {0,1,2 \ldots m - 1} \right\}\) , \(m \in \mathbb{N}\) , \(m \ge 3\) and \({ + _m}\) </span><span style="font-family: times new roman,times; font-size: medium;">denotes addition modulo \(m\) .</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 \(f\) is a bijection if \(\boldsymbol{A}\) is non-singular.</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">A.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;">Suppose now that \(\boldsymbol{A}\) is singular.</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) Write down the relationship between \(a\) , \(b\) , \(c\) , \(d\) .</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (ii) Deduce that the second row of \(\boldsymbol{A}\) is a multiple of the first row of \(\boldsymbol{A}\) .</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (iii) Hence show that \(f\) is not a bijection.</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">A.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;">Show that \(\left\{ {S,{ + _m}} \right\}\) is cyclic for all <strong><em>m</em></strong> .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">B.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;">Given that \(m\) is prime,</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) explain why all elements except the identity are generators of \(\left\{ {S,{ + _m}} \right\}\) ;</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (ii) find the inverse of \(x\) , where <strong><em>x</em></strong> is any element of \(\left\{ {S,{ + _m}} \right\}\) apart from the </span><span style="font-family: times new roman,times; font-size: medium;">identity;</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (iii) determine the number of sets of two distinct elements where each element </span><span style="font-family: times new roman,times; font-size: medium;">is the inverse of the other.</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">B.b.</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;">Suppose now that \(m = ab\) where \(a\) , \(b\) are unequal prime numbers. Show that </span><span style="font-family: times new roman,times; font-size: medium;">\(\left\{ {S,{ + _m}} \right\}\) has two proper subgroups and identify them.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">B.c.</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;">recognizing that the function needs to be injective and surjective <em><strong>R1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>R1</strong></em> if this is seen anywhere in the solution.</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;">injective:</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">let \(\boldsymbol{U}, \boldsymbol{V} \in ^\circ \times ^\circ \) be 2-D column vectors such that \(\boldsymbol{AU} = \boldsymbol{AV}\) <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({\boldsymbol{A}^{ - 1}}\boldsymbol{AU} = {\boldsymbol{A}^{ - 1}}\boldsymbol{AV}\) <strong><em>M1</em></strong></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\boldsymbol{U} = \boldsymbol{V}\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">this shows that \(f\) is injective</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">surjective:</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">let \(W \in ^\circ \times ^\circ \) <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">then there exists \(\boldsymbol{Z} = {\boldsymbol{A}^{ - 1}}\boldsymbol{W} \in ^\circ \times ^\circ \) such that \(\boldsymbol{AZ} = \boldsymbol{W}\) <strong><em>M1A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">this shows that \(f\) is surjective</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">therefore \(f\) is a bijection <em><strong>AG</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[7 marks]</span></strong></em></p>
<div class="question_part_label">A.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) the relationship is \(ad = bc\) <em><strong>A1</strong></em></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) it follows that \(\frac{c}{a} = \frac{d}{b} = \lambda \) </span><span style="font-family: times new roman,times; font-size: medium;">so that \((c,d) = \lambda (a,b)\) <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;">(iii) <strong>EITHER</strong></span></p>
<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;">\(\boldsymbol{W} = \left[ \begin{array}{l}<br>p\\<br>q<br>\end{array} \right]\) be a 2-D vector</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">then </span><span style="font-family: times new roman,times; font-size: medium;">\(\boldsymbol{AW} = \left[ \begin{array}{l}<br>a\\<br>\lambda a<br>\end{array} \right.\left. \begin{array}{l}<br>b\\<br>\lambda b<br>\end{array} \right]\left[ \begin{array}{l}<br>p\\<br>q<br>\end{array} \right]\) <strong><em>M1</em></strong></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = \left[ \begin{array}{l}<br>ap + bq\\<br>\lambda (ap + bq)<br>\end{array} \right]\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the image always satisfies \(y = \lambda x\) so \(f\) is not surjective and therefore </span><span style="font-family: times new roman,times; font-size: medium;">not a bijection <strong><em>R1</em></strong></span></p>
<p align="LEFT"><strong><span style="font-family: times new roman,times; font-size: medium;">OR</span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">consider</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left[ {\begin{array}{*{20}{c}}<br> a&b \\ <br> {\lambda a}&{\lambda b} <br>\end{array}} \right]\left[ {\begin{array}{*{20}{c}}<br> b \\ <br> 0 <br>\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}<br> {ab} \\ <br> {\lambda ab} <br>\end{array}} \right]\)<br></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left[ {\begin{array}{*{20}{c}}<br> a&b \\ <br> {\lambda a}&{\lambda b} <br>\end{array}} \right]\left[ {\begin{array}{*{20}{c}}<br> 0 \\ <br> a <br>\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}<br> {ab} \\ <br> {\lambda ab} <br>\end{array}} \right]\)<br></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">this shows that \(f\) is not injective and therefore not a bijection <strong><em>R1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> </span></em></strong></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.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the identity element is \(0\) <strong><em> R1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">consider, for \(1 \le r \le m\) ,</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">using \(1\) as a generator <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(1\) combined with itself \(r\) times gives \(r\) and as \(r\) increases from \(1\) to m, the </span><span style="font-family: times new roman,times; font-size: medium;">group is generated ending with \(0\) when \(r = m\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">it is therefore cyclic <strong><em>AG</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">B.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) by Lagrange the order of each element must be a factor of \(m\) and if \(m\) </span><span style="font-family: times new roman,times; font-size: medium;">is prime, its only factors are \(1\) and \(m\) <strong><em>R1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">since 0 is the only element of order \(1\), all other elements are of order \(m\) </span><span style="font-family: times new roman,times; font-size: medium;">and are therefore generators <strong><em>R1</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) since \(x{ + _m}(m - x) = 0\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">(M1)</span></em></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the inverse of <strong><em>x</em></strong> is \((m - x)\) <strong><em>A1</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;">(iii) consider</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/america.png" alt> </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1A1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">there are \(\frac{1}{2}(m - 1)\) </span><span style="font-family: times new roman,times; font-size: medium;">inverse pairs <strong><em>A1 N1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <strong><em>M1</em></strong> for an attempt to list the inverse pairs, <strong><em>A1</em></strong> for completing it </span><span style="font-family: times new roman,times; font-size: medium;">correctly and <strong><em>A1</em></strong> for the final answer.</span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> </span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[7 marks]</span></strong></em></p>
<div class="question_part_label">B.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 \(a\), \(b\) are unequal primes the only factors of \(m\) are \(a\) and \(b\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">there are therefore only subgroups of order \(a\) and \(b\) <em><strong> R1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">they are</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\left\{ {0,a,2a, \ldots ,(b - 1)a} \right\}\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\left\{ {0,b,2b, \ldots ,(a - 1)b} \right\}\) <strong><em>A1</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">B.c.</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 proved to be a difficult question for some candidates. Most candidates realised that they had to show that the function was both injective and surjective but many failed to give convincing proofs. Some candidates stated, incorrectly, that <strong><em>f</em></strong> was injective because \(\boldsymbol{AX}\) is uniquely defined, not realising that they had to show that \(\boldsymbol{AX} = \boldsymbol{AY} \Rightarrow \boldsymbol{X} = \boldsymbol{Y}\) . </span></p>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Solutions to (b) were disappointing with many candidates failing to realise that they had either to show that \(\boldsymbol{AX}\) was confined to a subset of \(\mathbb{R} \times \mathbb{R}\)</span><span style="font-family: times new roman,times; font-size: medium;"> or that two distinct vectors had the same image under \(f\). </span></p>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question was well answered in general with solutions to (c) being the least successful. </span></p>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question was well answered in general with solutions to (c) being the least successful. </span></p>
<div class="question_part_label">B.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question was well answered in general with solutions to (c) being the least successful. </span></p>
<div class="question_part_label">B.c.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">By considering the points \((1,{\text{ }}0)\) and \((0,{\text{ }}1)\) determine the \(2 \times 2\) <span class="s1">matrix which represents</span></p>
<p class="p2">(i) <span class="Apple-converted-space"> </span>an anticlockwise rotation of \(\theta \) about the origin;</p>
<p class="p2">(ii) <span class="Apple-converted-space"> </span>a reflection in the line \(y = (\tan \theta )x\).</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 class="p1">Determine the matrix \(A\) which represents a rotation from the direction \(\left( {\begin{array}{*{20}{c}} 1 \\ 0 \end{array}} \right)\) to the direction \(\left( {\begin{array}{*{20}{c}} 1 \\ 3 \end{array}} \right)\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1"><span class="s1">A triangle whose vertices have coordinates \((0,{\text{ }}0)\), \((3,{\text{ }}1)\) and \((1,{\text{ }}5)\) </span>undergoes a transformation represented by the matrix \({A^{ - 1}}XA\), where \(X\) is the matrix representing a reflection in the <span class="s1">\(x\)-axis</span>. Find the coordinates of the vertices of the transformed triangle.</p>
<div class="marks">[6]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">The matrix \(B = {A^{ - 1}}XA\) represents a reflection in the line \(y = mx\). Find the value of \(m\).</p>
<div class="marks">[6]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>(i) under an anti-clockwise rotation of \(\theta \)</p>
<p>\((1,{\text{ }}0) \to (\cos \theta ,{\text{ }}\sin \theta )\)</p>
<p>\((0,{\text{ }}1) \to ( - \sin \theta ,{\text{ }}\cos \theta )\) <strong><em>M1</em></strong></p>
<p>rotation matrix is \(\left( {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta } \\ {\sin \theta }&{\cos \theta } \end{array}} \right)\) <strong><em>A1</em></strong></p>
<p> </p>
<p>(ii) <img src="data:image/png;base64,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" alt> <strong><em>M1</em></strong></p>
<p>under a reflection in the line \(y = (\tan \theta )x\)</p>
<p>\((1,{\text{ }}0) \to (\cos 2\theta ,{\text{ }}\sin 2\theta )\)</p>
<p>\((0,{\text{ }}1) \to (\sin 2\theta ,{\text{ }} - \cos 2\theta )\) <strong><em>M1</em></strong></p>
<p>matrix for reflection in the line \(y = (\tan \theta )x: \left( {\begin{array}{*{20}{c}} {\cos 2\theta }&{\sin 2\theta } \\ {\sin 2\theta }&{ - \cos 2\theta } \end{array}} \right)\) <strong><em>A1</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">in this case \(\tan \theta = 3\) <span class="Apple-converted-space"> </span>(<strong><em>M1)</em></strong></p>
<p class="p1">\(\Rightarrow \sin \theta = \frac{3}{{\sqrt {10} }}\)</p>
<p class="p1">hence rotation matrix is \(\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt {10} }}}&{ - \frac{3}{{\sqrt {10} }}} \\ {\frac{3}{{\sqrt {10} }}}&{\frac{1}{{\sqrt {10} }}} \end{array}} \right)\) <strong><em>A1</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">\({A^{ - 1}} = \left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt {10} }}}&{\frac{3}{{\sqrt {10} }}} \\ { - \frac{3}{{\sqrt {10} }}}&{\frac{1}{{\sqrt {10} }}} \end{array}} \right)\) <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1">\(X = \left( {\begin{array}{*{20}{c}} 1&0 \\ 0&{ - 1} \end{array}} \right)\) <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1">\( \Rightarrow {A^{ - 1}}XA = \left( {\begin{array}{*{20}{c}} { - \frac{8}{{10}}}&{ - \frac{6}{{10}}} \\ { - \frac{6}{{10}}}&{\frac{8}{{10}}} \end{array}} \right)\) <span class="Apple-converted-space"> </span><strong><em>(M1)A1</em></strong></p>
<p class="p1">\( \Rightarrow {A^{ - 1}}XA(G) = \left( {\begin{array}{*{20}{c}} { - \frac{8}{{10}}}&{ - \frac{6}{{10}}} \\ { - \frac{6}{{10}}}&{\frac{8}{{10}}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} 0&3&1 \\ 0&1&5 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0&{ - 3}&{ - \frac{{19}}{5}} \\ 0&{ - 1}&{\frac{{17}}{5}} \end{array}} \right)\) <span class="Apple-converted-space"> </span>(<strong><em>M1)</em></strong></p>
<p class="p1">hence coordinates are \((0,{\text{ }}0)\), \(( - 3,{\text{ }} - 1)\) and \(\left( { - \frac{{19}}{5},{\text{ }}\frac{{17}}{5}} \right)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">\(B = \left( {\begin{array}{*{20}{c}} { - \frac{8}{{10}}}&{ - \frac{6}{{10}}} \\ { - \frac{6}{{10}}}&{\frac{8}{{10}}} \end{array}} \right)\)</p>
<p class="p1">the matrix for the reflection in the line \(y = (\tan \theta )x\) is \(\left( {\begin{array}{*{20}{c}} {\cos 2\theta }&{\sin 2\theta } \\ {\sin 2\theta }&{ - \cos 2\theta } \end{array}} \right)\)</p>
<p class="p1">\(\cos 2\theta = - \frac{4}{5},{\text{ }}\sin 2\theta = - \frac{3}{5}\) (<strong><em>A1)(A1)</em></strong></p>
<p class="p1">\(\cos 2\theta = 2{\cos ^2}\theta - 1\) <strong><em>(M1)</em></strong></p>
<p class="p1">\( \Rightarrow 2{\cos ^2}\theta = \frac{2}{{10}}\)</p>
<p class="p1">\( \Rightarrow \cos \theta = \pm \frac{1}{{\sqrt {10} }}\) (<strong><em>A1)</em></strong></p>
<p class="p1">\( \Rightarrow \cos \theta = - \frac{1}{{\sqrt {10} }}\) and \(\sin \theta = \frac{3}{{\sqrt {10} }}\) (<strong><em>A1)</em></strong></p>
<p class="p1">\( \Rightarrow \tan \theta = - 3\)</p>
<p class="p1">\( \Rightarrow m = - 3\) <strong><em>A1</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">This proved to be a more challenging question for many candidates. In part a) many candidates appeared to not know how to find the matrices and for those who attempted to find them, arithmetic errors were common. A number of wholly correct solutions to parts b), c) and d) were seen, but many candidates seemed unfamiliar with this style of question and made errors or simply gave up part way through the process.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">This proved to be a more challenging question for many candidates. In part a) many candidates appeared to not know how to find the matrices and for those who attempted to find them, arithmetic errors were common. A number of wholly correct solutions to parts b), c) and d) were seen, but many candidates seemed unfamiliar with this style of question and made errors or simply gave up part way through the process.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">This proved to be a more challenging question for many candidates. In part a) many candidates appeared to not know how to find the matrices and for those who attempted to find them, arithmetic errors were common. A number of wholly correct solutions to parts b), c) and d) were seen, but many candidates seemed unfamiliar with this style of question and made errors or simply gave up part way through the process.</p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">This proved to be a more challenging question for many candidates. In part a) many candidates appeared to not know how to find the matrices and for those who attempted to find them, arithmetic errors were common. A number of wholly correct solutions to parts b), c) and d) were seen, but many candidates seemed unfamiliar with this style of question and made errors or simply gave up part way through the process.</p>
<div class="question_part_label">d.</div>
</div>
<br><hr><br>