7 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { M }\), where $$\mathbf { M } = \left( \begin{array} { r r r r } 1 & - 2 & - 3 & 1 \\ 3 & - 5 & - 7 & 7 \\ 5 & - 9 & - 13 & 9 \\ 7 & - 13 & - 19 & 11 \end{array} \right)$$ Find the rank of \(\mathbf { M }\) and a basis for the null space of T . The vector \(\left( \begin{array} { l } 1 \\ 2 \\ 3 \\ 4 \end{array} \right)\) is denoted by \(\mathbf { e }\). Show that there is a solution of the equation \(\mathbf { M x } = \mathbf { M e }\) of the form \(\mathbf { x } = \left( \begin{array} { c } a \\ b \\ - 1 \\ - 1 \end{array} \right)\), where the constants \(a\) and \(b\) are to be found.

7 The linear transformation $\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }$ is represented by the matrix $\mathbf { M }$, where

$$\mathbf { M } = \left( \begin{array} { r r r r } 
1 & - 2 & - 3 & 1 \\
3 & - 5 & - 7 & 7 \\
5 & - 9 & - 13 & 9 \\
7 & - 13 & - 19 & 11
\end{array} \right)$$

Find the rank of $\mathbf { M }$ and a basis for the null space of T .

The vector $\left( \begin{array} { l } 1 \\ 2 \\ 3 \\ 4 \end{array} \right)$ is denoted by $\mathbf { e }$. Show that there is a solution of the equation $\mathbf { M x } = \mathbf { M e }$ of the form $\mathbf { x } = \left( \begin{array} { c } a \\ b \\ - 1 \\ - 1 \end{array} \right)$, where the constants $a$ and $b$ are to be found.

\hfill \mbox{\textit{CAIE FP1 2015 Q7}}