Question 7 - A-Level Maths

CAIE FP1 2015 November — Question 7

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2015
Session	November
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Matrices
Type	Solving linear systems using matrices
Difficulty	Challenging +1.2 This is a systematic Further Maths linear algebra question requiring row reduction to find rank and null space, followed by solving a matrix equation with a specified form. While it involves multiple steps and FM content (rank-nullity, null space basis), the techniques are standard algorithmic procedures taught directly in FP1 with no novel insight required. The specified form in part (ii) actually simplifies the work. Moderately above average due to FM content and computational length, but straightforward application of taught methods.
Spec	4.03a Matrix language: terminology and notation 4.03r Solve simultaneous equations: using inverse matrix

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.

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Question 7:

Answer	Marks	Guidance
Working/Answer	Marks	Guidance
$\begin{pmatrix}1&-2&-3&1\\3&-5&-7&7\\5&-9&-13&9\\7&-13&-19&11\end{pmatrix} \sim \begin{pmatrix}1&-2&-3&1\\0&1&2&4\\0&1&2&4\\0&1&2&4\end{pmatrix} \sim \begin{pmatrix}1&-2&-3&1\\0&1&2&4\\0&0&0&0\\0&0&0&0\end{pmatrix}$	M1A1
$r(\mathbf{M}) = 4 - 2 = 2$	A1
$x - 2y - 3z + t = 0$ and $y + 2z + 4t = 0$	M1
Set $z = \lambda$ and $t = \mu \Rightarrow y = -2\lambda - 4\mu$ and $x = -\lambda - 9\mu$	M1
Basis is $\left\{\begin{pmatrix}-1\\-2\\1\\0\end{pmatrix}, \begin{pmatrix}-9\\-4\\0\\1\end{pmatrix}\right\}$	A1
$\mathbf{x} = \begin{pmatrix}1\\2\\3\\4\end{pmatrix} + \lambda\begin{pmatrix}-1\\-2\\1\\0\end{pmatrix} + \mu\begin{pmatrix}-9\\-4\\0\\1\end{pmatrix} = \begin{pmatrix}a\\b\\-1\\-1\end{pmatrix}$	M1
Solving: $\lambda = -4$ and $\mu = -5$	M1A1
$\Rightarrow a = 50$, $b = 30$	A1
Total: 10 marks	[10]

## Question 7:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix}1&-2&-3&1\\3&-5&-7&7\\5&-9&-13&9\\7&-13&-19&11\end{pmatrix} \sim \begin{pmatrix}1&-2&-3&1\\0&1&2&4\\0&1&2&4\\0&1&2&4\end{pmatrix} \sim \begin{pmatrix}1&-2&-3&1\\0&1&2&4\\0&0&0&0\\0&0&0&0\end{pmatrix}$ | M1A1 | |
| $r(\mathbf{M}) = 4 - 2 = 2$ | A1 | |
| $x - 2y - 3z + t = 0$ and $y + 2z + 4t = 0$ | M1 | |
| Set $z = \lambda$ and $t = \mu \Rightarrow y = -2\lambda - 4\mu$ and $x = -\lambda - 9\mu$ | M1 | |
| Basis is $\left\{\begin{pmatrix}-1\\-2\\1\\0\end{pmatrix}, \begin{pmatrix}-9\\-4\\0\\1\end{pmatrix}\right\}$ | A1 | |
| $\mathbf{x} = \begin{pmatrix}1\\2\\3\\4\end{pmatrix} + \lambda\begin{pmatrix}-1\\-2\\1\\0\end{pmatrix} + \mu\begin{pmatrix}-9\\-4\\0\\1\end{pmatrix} = \begin{pmatrix}a\\b\\-1\\-1\end{pmatrix}$ | M1 | |
| Solving: $\lambda = -4$ and $\mu = -5$ | M1A1 | |
| $\Rightarrow a = 50$, $b = 30$ | A1 | |
| **Total: 10 marks** | [10] | |

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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}}

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