Question 3 - A-Level Maths

Edexcel CP2 2023 June — Question 3 10 marks

Exam Board	Edexcel
Module	CP2 (Core Pure 2)
Year	2023
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Invariant lines and eigenvalues and vectors
Type	Find invariant lines through origin
Difficulty	Standard +0.3 This is a structured multi-part question on matrix transformations requiring matrix multiplication, solving simultaneous equations from matrix equality, finding eigenvalues/eigenvectors, and interpreting invariant lines. While it involves several steps, each part follows standard procedures taught in Core Pure 2 with clear scaffolding, making it slightly easier than average A-level difficulty.
Spec	4.03a Matrix language: terminology and notation 4.03b Matrix operations: addition, multiplication, scalar 4.03g Invariant points and lines

3. $$\mathbf { M } = \left( \begin{array} { r r } - 2 & 5 \\ 6 & k \end{array} \right)$$ where $k$ is a constant.
Given that $$\mathbf { M } ^ { 2 } + 11 \mathbf { M } = a \mathbf { I }$$ where $a$ is a constant and $\mathbf { I }$ is the $2 \times 2$ identity matrix,

1. determine the value of $a$
2. show that $k = - 9$
Determine the equations of the invariant lines of the transformation represented by $\mathbf { M }$.
State which, if any, of the lines identified in (b) consist of fixed points, giving a reason for your answer.

Show mark scheme Show mark scheme source

Question 3:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\mathbf{M}^2+11\mathbf{M} = \begin{pmatrix}a&0\\0&a\end{pmatrix} \Rightarrow \begin{pmatrix}34&5k-10\\6k-12&k^2+30\end{pmatrix}+\begin{pmatrix}-22&55\\66&11k\end{pmatrix}=\begin{pmatrix}a&0\\0&a\end{pmatrix}$	M1	Evaluates $\mathbf{M}^2$ and uses in the equation given
$\Rightarrow a=12$	A1	Correct value of $a$ deduced
$5k-10+55=0 \Rightarrow k=-9$; or $6k-12+66=0 \Rightarrow k=-9$; or $k^2+11k+18=0 \Rightarrow k=-2,-9^*$, $k\neq -2$	A1\*	Correct work to show $k=-9$. If off-diagonals used no further justification needed. If bottom right entry used must have valid reason for rejecting $-2$
(3)

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\begin{pmatrix}-2&5\\6&-9\end{pmatrix}\begin{pmatrix}x\\mx+c\end{pmatrix}=\begin{pmatrix}X\\mX+c\end{pmatrix}\Rightarrow\begin{cases}-2x+5(mx+c)=X\\6x-9(mx+c)=mX+c\end{cases}$	M1	Sets up the matrix equation for invariant lines and extracts the simultaneous equations
$\Rightarrow 6x-9mx-9c=-2mx+5m^2x+5mc+\ldots$	M1	Eliminates $X$ to get a linear equation in $x$
$\Rightarrow(5m^2+7m-6)x+(5m+10)c=0$	A1	Correct equation, need not be simplified
$\Rightarrow 5m^2+7m-6=0\Rightarrow(m+2)(5m-3)\Rightarrow m=-2,\frac{3}{5}$	M1	Solves their quadratic in $m$ by any valid means including calculator
$m=\frac{3}{5}\Rightarrow 5m+10\neq 0$ so need $c=0$, hence $y=\frac{3}{5}x$ is a fixed line	A1	Deduces $y=\frac{3}{5}x$ is a fixed line (where $c=0$)
$m=-2\Rightarrow 5m+10=0$ so $c$ can be anything, so $y=-2x+c$ for any $c$ is fixed	A1	Deduces $y=-2x+c$ is a fixed line where $c$ can be any value. Must include all lines
(6)

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$(0,c)\to(5c,-9c)$ so need $c=0$; $(1,m)\to(-2+5m, 6-9m)$ so need $5m=3$, hence $y=\frac{3}{5}x$ contains fixed points	B1
(1)
(10 marks total)

Question 3 (Matrix/Fixed Points):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\begin{pmatrix}-2 & 5\\ 6 & 9\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} = \begin{pmatrix}x\\ y\end{pmatrix} \Rightarrow \{5y = 3x\}$	A1	One correct equation
Uses both eigenvalues to find an equation	M1	Uses both of their eigenvalues to find an equation
Deduces $y = \frac{3}{5}x$	A1	—
Deduces $y = -2x + c$	A1	—

(c)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$y = \frac{3}{5}x$ is a line of fixed points	B1	Identifies $y = \frac{3}{5}x$ is a line of fixed points with reason. Allow if $c = 0$ is assumed.

## Question 3:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{M}^2+11\mathbf{M} = \begin{pmatrix}a&0\\0&a\end{pmatrix} \Rightarrow \begin{pmatrix}34&5k-10\\6k-12&k^2+30\end{pmatrix}+\begin{pmatrix}-22&55\\66&11k\end{pmatrix}=\begin{pmatrix}a&0\\0&a\end{pmatrix}$ | **M1** | Evaluates $\mathbf{M}^2$ and uses in the equation given |
| $\Rightarrow a=12$ | **A1** | Correct value of $a$ deduced |
| $5k-10+55=0 \Rightarrow k=-9$; or $6k-12+66=0 \Rightarrow k=-9$; or $k^2+11k+18=0 \Rightarrow k=-2,-9^*$, $k\neq -2$ | **A1\*** | Correct work to show $k=-9$. If off-diagonals used no further justification needed. If bottom right entry used must have valid reason for rejecting $-2$ |
| | **(3)** | |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}-2&5\\6&-9\end{pmatrix}\begin{pmatrix}x\\mx+c\end{pmatrix}=\begin{pmatrix}X\\mX+c\end{pmatrix}\Rightarrow\begin{cases}-2x+5(mx+c)=X\\6x-9(mx+c)=mX+c\end{cases}$ | **M1** | Sets up the matrix equation for invariant lines and extracts the simultaneous equations |
| $\Rightarrow 6x-9mx-9c=-2mx+5m^2x+5mc+\ldots$ | **M1** | Eliminates $X$ to get a linear equation in $x$ |
| $\Rightarrow(5m^2+7m-6)x+(5m+10)c=0$ | **A1** | Correct equation, need not be simplified |
| $\Rightarrow 5m^2+7m-6=0\Rightarrow(m+2)(5m-3)\Rightarrow m=-2,\frac{3}{5}$ | **M1** | Solves their quadratic in $m$ by any valid means including calculator |
| $m=\frac{3}{5}\Rightarrow 5m+10\neq 0$ so need $c=0$, hence $y=\frac{3}{5}x$ is a fixed line | **A1** | Deduces $y=\frac{3}{5}x$ is a fixed line (where $c=0$) |
| $m=-2\Rightarrow 5m+10=0$ so $c$ can be anything, so $y=-2x+c$ for any $c$ is fixed | **A1** | Deduces $y=-2x+c$ is a fixed line where $c$ can be any value. Must include all lines |
| | **(6)** | |

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(0,c)\to(5c,-9c)$ so need $c=0$; $(1,m)\to(-2+5m, 6-9m)$ so need $5m=3$, hence $y=\frac{3}{5}x$ contains fixed points | **B1** | |
| | **(1)** | |
| | **(10 marks total)** | |

# Question 3 (Matrix/Fixed Points):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}-2 & 5\\ 6 & 9\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} = \begin{pmatrix}x\\ y\end{pmatrix} \Rightarrow \{5y = 3x\}$ | A1 | One correct equation |
| Uses both eigenvalues to find an equation | M1 | Uses both of their eigenvalues to find an equation |
| Deduces $y = \frac{3}{5}x$ | A1 | — |
| Deduces $y = -2x + c$ | A1 | — |

**(c)**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{3}{5}x$ is a line of fixed points | B1 | Identifies $y = \frac{3}{5}x$ is a line of fixed points with reason. Allow if $c = 0$ is assumed. |

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Show LaTeX source

3.

$$\mathbf { M } = \left( \begin{array} { r r } 
- 2 & 5 \\
6 & k
\end{array} \right)$$

where $k$ is a constant.\\
Given that

$$\mathbf { M } ^ { 2 } + 11 \mathbf { M } = a \mathbf { I }$$

where $a$ is a constant and $\mathbf { I }$ is the $2 \times 2$ identity matrix,
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item determine the value of $a$
\item show that $k = - 9$
\end{enumerate}\item Determine the equations of the invariant lines of the transformation represented by $\mathbf { M }$.
\item State which, if any, of the lines identified in (b) consist of fixed points, giving a reason for your answer.
\end{enumerate}

\hfill \mbox{\textit{Edexcel CP2 2023 Q3 [10]}}

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