The linear transformation $\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }$ is represented by the matrix $\mathbf { M } = \left( \begin{array} { r r r r } 1 & 1 & 5 & 7 \\ 3 & 9 & 17 & 25 \\ 1 & 7 & 7 & 11 \\ 3 & 6 & 16 & 23 \end{array} \right)$.\\
(i) In either order,
\begin{enumerate}[label=(\alph*)]
\item show that the dimension of $R$, the range space of T , is equal to 2 ,
\item obtain a basis for $R$.\\
(ii) Show that the vector $\left( \begin{array} { r } 1 \\ - 15 \\ - 17 \\ - 6 \end{array} \right)$ belongs to $R$.\\
(iii) It is given that $\left\{ \mathbf { e } _ { 1 } , \mathbf { e } _ { 2 } \right\}$ is a basis for the null space of T , where $\mathbf { e } _ { 1 } = \left( \begin{array} { r } 14 \\ 1 \\ - 3 \\ 0 \end{array} \right)$ and $\mathbf { e } _ { 2 } = \left( \begin{array} { r } 19 \\ 2 \\ 0 \\ - 3 \end{array} \right)$. Show that, for all $\lambda$ and $\mu$,

$$\mathbf { x } = \left( \begin{array} { r } 
4 \\
- 3 \\
0 \\
0
\end{array} \right) + \lambda \mathbf { e } _ { 1 } + \mu \mathbf { e } _ { 2 }$$

is a solution of

$$\mathbf { M x } = \left( \begin{array} { r } 
1 \\
- 15 \\
- 17 \\
- 6
\end{array} \right)$$

(iv) Hence find a solution of $( * )$ of the form $\left( \begin{array} { c } \alpha \\ 0 \\ \gamma \\ \delta \end{array} \right)$.
\end{enumerate}

\hfill \mbox{\textit{CAIE FP1 2010 Q12 OR}}