6 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { M }\), where $$\mathbf { M } = \left( \begin{array} { r r r r } 1 & - 3 & - 1 & 2
4 & - 10 & 0 & 2
1 & - 1 & 3 & - 4
5 & - 12 & 1 & 1 \end{array} \right)$$ Find, in either order, the rank of \(\mathbf { M }\) and a basis for the null space \(K\) of T. Evaluate $$\mathbf { M } \left( \begin{array} { r } 1
- 2
- 3
- 4 \end{array} \right)$$ and hence show that every solution of $$\mathbf { M x } = \left( \begin{array} { r } 2
16
10
22 \end{array} \right)$$ has the form $$\mathbf { x } = \left( \begin{array} { r } 1
- 2
- 3
- 4 \end{array} \right) + \lambda \mathbf { e } _ { 1 } + \mu \mathbf { e } _ { 2 }$$ where \(\lambda\) and \(\mu\) are real numbers and \(\left\{ \mathbf { e } _ { 1 } , \mathbf { e } _ { 2 } \right\}\) is a basis for \(K\).