Challenging +1.2 This is a standard Further Maths linear algebra question requiring row reduction to find null space bases for two parameter values. While it involves 4×4 matrices and requires systematic Gaussian elimination twice, the technique is routine for FP1 students and follows a well-practiced algorithm with no conceptual surprises or novel problem-solving required.
6 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { M }\), where
$$\mathbf { M } = \left( \begin{array} { r r r r }
- 2 & 5 & 3 & - 1 \\
0 & 1 & - 4 & - 2 \\
6 & - 14 & - 13 & 1 \\
\alpha & \alpha & - 2 \alpha & - 11 \alpha
\end{array} \right)$$
and \(\alpha\) is a constant. The null space of T is denoted by \(K _ { 1 }\) when \(\alpha \neq 0\), and by \(K _ { 2 }\) when \(\alpha = 0\). Find a basis for \(K _ { 1 }\) and a basis for \(K _ { 2 }\).
\(\Rightarrow K_2\left\{\begin{pmatrix}23\\8\\2\\0\end{pmatrix}, \begin{pmatrix}9\\4\\0\\2\end{pmatrix}\right\}\) (OE) e.g. \(\begin{pmatrix}5\\0\\2\\-4\end{pmatrix}\) or \(\begin{pmatrix}0\\10\\9\\-23\end{pmatrix}\)
A1A1
8 marks total
## Question 6:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix}-2&5&3&-1\\0&1&-4&-2\\6&-14&-13&1\\1&1&-2&-11\end{pmatrix} \sim \begin{pmatrix}-2&5&3&-1\\0&1&-4&-2\\0&1&-4&-2\\0&7&-1&-23\end{pmatrix}$ | M1A1 | Reduces to echelon form |
| $\sim \begin{pmatrix}-2&5&3&-1\\0&1&-4&-2\\0&0&0&0\\0&0&3&-1\end{pmatrix} \sim \begin{pmatrix}-2&5&3&-1\\0&1&-4&-2\\0&0&3&-1\\0&0&0&0\end{pmatrix}\ (\alpha\neq 0)$ | A1 | N.B. Allow matrix with row of zeros not in echelon form |
| $-2x+5y+3z-t=0$; $y-4z-2t=0$; $3z-t=0$ | | |
| $\Rightarrow K_1\left\{\begin{pmatrix}25\\10\\1\\3\end{pmatrix}\right\}$ (OE) | M1 A1 | Solves set of equations; obtains basis |
| If $\alpha=0$: $-2x+5y+3z-t=0$; $y-4z-2t=0$ | M1 | Solves equations in second case |
| $\Rightarrow K_2\left\{\begin{pmatrix}23\\8\\2\\0\end{pmatrix}, \begin{pmatrix}9\\4\\0\\2\end{pmatrix}\right\}$ (OE) e.g. $\begin{pmatrix}5\\0\\2\\-4\end{pmatrix}$ or $\begin{pmatrix}0\\10\\9\\-23\end{pmatrix}$ | A1A1 | 8 marks total |
6 The linear transformation $\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }$ is represented by the matrix $\mathbf { M }$, where
$$\mathbf { M } = \left( \begin{array} { r r r r }
- 2 & 5 & 3 & - 1 \\
0 & 1 & - 4 & - 2 \\
6 & - 14 & - 13 & 1 \\
\alpha & \alpha & - 2 \alpha & - 11 \alpha
\end{array} \right)$$
and $\alpha$ is a constant. The null space of T is denoted by $K _ { 1 }$ when $\alpha \neq 0$, and by $K _ { 2 }$ when $\alpha = 0$. Find a basis for $K _ { 1 }$ and a basis for $K _ { 2 }$.
\hfill \mbox{\textit{CAIE FP1 2013 Q6 [8]}}