The matrix A is given by A = $\left( {\begin{array}{*{20}{c}}1&2&3&4\\3&8&{11}&8\\1&3&4&\lambda \\\lambda &5&7&6\end{array}} \right)$.

(a)     Given that $\lambda  = 2$, B = $\left( \begin{array}{l}2\\4\\\mu \\3\end{array} \right)$ and X = $\left( \begin{array}{l}x\\y\\z\\t\end{array} \right)$,

(i)     find the value of $\mu$ for which the equations defined by AX = B are consistent and solve the equations in this case;

(ii)     define the rank of a matrix and state the rank of A.

(b)     Given that $\lambda  = 1$,

(i)     show that the four column vectors in A form a basis for the space of four-dimensional column vectors;

(ii)     express the vector $\left( \begin{array}{c}6\\28\\12\\15\end{array} \right)$ as a linear combination of these basis vectors.

(a)     (i)     using row reduction,     M1

$\left( {\begin{array}{*{20}{c}}1&2&3&4\\3&8&{11}&8\\1&3&4&2\\2&5&7&6\end{array}{\rm{\,\,\,\,\,\,\,\,}}\begin{array}{*{20}{c}}2\\4\\\mu \\3\end{array}} \right)$

$\left( {\begin{array}{*{20}{c}}1&2&3&4\\0&2&2&{ - 4}\\0&1&1&{ - 2}\\0&1&1&{ - 2}\end{array}{\rm{\,\,\,\,\,\,\,\,}}\begin{array}{*{20}{c}}2\\{ - 2}\\{\mu  - 2}\\{ - 1}\end{array}} \right)$     (A2)

for consistency,

$\mu  - 2 =  - 1$     (M1)

$\mu  = 1$     A1

put $z = \alpha ,{\text{ }}t = \beta$     M1

$y =  - 1 - \alpha  + 2\beta ;{\text{ }}x = 4 - \alpha  - 8\beta$     A1A1

(ii)     the rank of a matrix is the number of independent rows (or columns)     A1

${\text{rank}}\,{\text{(}}$A$) = 2$     A1

[10 marks]

(b)     (i)     ${\text{det}}\,{\text{(}}$A$) = 2$     (M1)A1

since ${\text{det}}\,{\text{(}}$A$) \ne 0$, the vectors form a basis     R1

(ii)     let $\left( \begin{array}{l}6\\28\\12\\15\end{array} \right) = a\left( \begin{array}{l}1\\3\\1\\1\end{array} \right) + b\left( \begin{array}{l}2\\8\\3\\5\end{array} \right) + c\left( \begin{array}{c}3\\11\\4\\7\end{array} \right) + d\left( \begin{array}{l}4\\8\\1\\6\end{array} \right)$     M1

$= \left( {\begin{array}{*{20}{c}}1&2&3&4\\3&8&{11}&8\\1&3&4&1\\1&5&7&6\end{array}} \right)\left( \begin{array}{l}a\\b\\c\\d\end{array} \right)$

it follows that

$\left( \begin{array}{l}a\\b\\c\\d\end{array} \right) = {\left( {\begin{array}{*{20}{c}}1&2&3&4\\3&8&{11}&8\\1&3&4&1\\1&5&7&6\end{array}} \right)^{ - 1}}\left( \begin{array}{l}6\\28\\12\\15\end{array} \right)$

$= \left( \begin{array}{c}2\\1\\2\\ - 1\end{array} \right)$

therefore

$a = 2$     A1

$b = 1$     A1

$c = 2$     A1

$d =  - 1$     A1

$\left( \begin{array}{l}6\\28\\12\\15\end{array} \right) = 2\left( \begin{array}{l}1\\3\\1\\1\end{array} \right) + \left( \begin{array}{l}2\\8\\3\\5\end{array} \right) + 2\left( \begin{array}{c}3\\11\\4\\7\end{array} \right) - \left( \begin{array}{l}4\\8\\1\\6\end{array} \right)$

[8 marks]