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]