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 \(K\).

## Question 6:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix}1&-3&-1&2\\4&-10&0&2\\1&-1&3&-4\\5&-12&1&1\end{pmatrix} \rightarrow\ldots\rightarrow \begin{pmatrix}1&-3&-1&2\\0&1&2&-3\\0&0&0&0\\0&0&0&0\end{pmatrix}$ | M1A1 | Reduces to echelon form |
| $r(\mathbf{M}) = 4-2 = 2$ | A1 | Obtains rank |
| $x-3y-z+2t=0,\quad y+2z-3t=0$ | M1 | |
| $\Rightarrow t=\mu,\; z=\lambda,\; y=3\mu-2\lambda,\; x=7\mu-5\lambda$ | A1 | |
| Basis is: $\left\{\begin{pmatrix}-5\\-2\\1\\0\end{pmatrix}, \begin{pmatrix}7\\3\\0\\1\end{pmatrix}\right\}$ (OE) | A1 | and basis for null space |
| $\mathbf{M}\begin{pmatrix}1\\-2\\-3\\-4\end{pmatrix} = \begin{pmatrix}2\\16\\10\\22\end{pmatrix}$ | B1 | Evaluates matrix product |
| $\mathbf{x} - \begin{pmatrix}1\\-2\\-3\\-4\end{pmatrix} \in K$ | M1 | Finds general solution of equations |
| $\mathbf{x} = \begin{pmatrix}1\\-2\\-3\\-4\end{pmatrix} + \lambda\mathbf{e}_1 + \mu\mathbf{e}_2$ (AG) | A1 | |

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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 $K$.

\hfill \mbox{\textit{CAIE FP1 2013 Q6 [9]}}