Question 2 - A-Level Maths

WJEC Further Unit 1 2022 June — Question 2 11 marks

Exam Board	WJEC
Module	Further Unit 1 (Further Unit 1)
Year	2022
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Matrices
Type	Solving matrix equations for unknown matrix
Difficulty	Standard +0.3 Part (a) requires finding the inverse of a 2×2 matrix and multiplying to solve AX=B, which is a standard Further Maths technique. Part (b)(i) uses the standard reflection matrix formula, and (b)(ii) applies the transformation to two points and finds a midpoint. All steps are routine applications of learned methods with no novel problem-solving required, making this slightly easier than average for Further Maths.
Spec	4.03d Linear transformations 2D: reflection, rotation, enlargement, shear 4.03o Inverse 3x3 matrix 4.03r Solve simultaneous equations: using inverse matrix

2. (a) The matrices $\mathbf { A }$ and $\mathbf { B }$ are defined by $$\mathbf { A } = \left( \begin{array} { c c } 3 & 4 \\ - 1 & - 2 \end{array} \right) , \quad \mathbf { B } = \binom { - 11 } { 7 }$$ Given that $\mathbf { A X } = \mathbf { B }$, find the matrix $\mathbf { X }$.
(b) (i) Find the $2 \times 2$ matrix, $\mathbf { T }$, which represents a reflection in the line $y = - 2 x$.
(ii) The images of the points $C ( 2,7 )$ and $D ( 3,1 )$, under $\mathbf { T }$, are $E$ and $F$ respectively. Find the coordinates of the midpoint of $E F$.

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Question 2:

Part a):

Method 1:

Answer	Marks	Guidance
Working	Mark	Guidance
Let $X = \begin{pmatrix} a \\ b \end{pmatrix}$, $\begin{pmatrix} 3 & 4 \\ -1 & -2 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} -11 \\ 7 \end{pmatrix}$, giving $3a+4b=-11$, $-a-2b=7$	M1, A1	Attempt to form 2 simultaneous equations
Solving: $a=3$ and $b=-5$, $X = \begin{pmatrix} 3 \\ -5 \end{pmatrix}$	M1, A1	Attempt to solve; must be in matrix form

Method 2:

Answer	Marks	Guidance
Working	Mark	Guidance
$\det A = (3 \times -2)-(4 \times -1) = -2$	(B1)	si
$A^{-1} = \frac{1}{-2}\begin{pmatrix} -2 & -4 \\ 1 & 3 \end{pmatrix}$	(B1)
$X = A^{-1}B = \frac{1}{-2}\begin{pmatrix} -2 & -4 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} -11 \\ 7 \end{pmatrix}$	(M1)
$X = \begin{pmatrix} 3 \\ -5 \end{pmatrix}$	(A1)	Must be in matrix form
[4]

Part b(i)):

Answer	Marks	Guidance
Working	Mark	Guidance
If reflection in $y = -2x$, then $\tan\theta = -2$, $\therefore \theta = \tan^{-1}(-2)$	B1	si
Reflection matrix: $\begin{pmatrix} -\frac{3}{5} & -\frac{4}{5} \\ -\frac{4}{5} & \frac{3}{5} \end{pmatrix}$	B2	B1 for 1 error (possibly repeated); If B2 then -1 for PA
[3]

Part b(ii)):

Method 1:

Answer	Marks	Guidance
Working	Mark	Guidance
$EF = \begin{pmatrix} -\frac{3}{5} & -\frac{4}{5} \\ -\frac{4}{5} & \frac{3}{5} \end{pmatrix}\begin{pmatrix} 2 & 3 \\ 7 & 1 \end{pmatrix}$	M1	FT their $T$; for attempt to multiply at least 1 point matrix
$EF = \begin{pmatrix} -\frac{34}{5} & -\frac{13}{5} \\ \frac{13}{5} & -\frac{9}{5} \end{pmatrix}$	A1, A1	Left column; Right column (may be seen as separate matrices)
Midpoint: $\left(-\frac{47}{10}, \frac{2}{5}\right)$	B1	oe, FT their $E$ and $F$

Method 2:

Answer	Marks	Guidance
Working	Mark	Guidance
Midpoint of $CD = \left(\frac{2+3}{2}, \frac{7+1}{2}\right) = \left(\frac{5}{2}, 4\right)$	(B1)	FT their $T$
$\begin{pmatrix} -\frac{3}{5} & -\frac{4}{5} \\ -\frac{4}{5} & \frac{3}{5} \end{pmatrix}\begin{pmatrix} \frac{5}{2} \\ 4 \end{pmatrix}$	(M1)	FT their midpoint
$= \begin{pmatrix} -\frac{47}{10} \\ \frac{2}{5} \end{pmatrix}$	(A1)
Midpoint of $EF$: $\left(-\frac{47}{10}, \frac{2}{5}\right)$	(A1)	oe
[4]
[11]

# Question 2:

## Part a):

**Method 1:**

| Working | Mark | Guidance |
|---------|------|----------|
| Let $X = \begin{pmatrix} a \\ b \end{pmatrix}$, $\begin{pmatrix} 3 & 4 \\ -1 & -2 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} -11 \\ 7 \end{pmatrix}$, giving $3a+4b=-11$, $-a-2b=7$ | M1, A1 | Attempt to form 2 simultaneous equations |
| Solving: $a=3$ and $b=-5$, $X = \begin{pmatrix} 3 \\ -5 \end{pmatrix}$ | M1, A1 | Attempt to solve; must be in matrix form |

**Method 2:**

| Working | Mark | Guidance |
|---------|------|----------|
| $\det A = (3 \times -2)-(4 \times -1) = -2$ | (B1) | si |
| $A^{-1} = \frac{1}{-2}\begin{pmatrix} -2 & -4 \\ 1 & 3 \end{pmatrix}$ | (B1) | |
| $X = A^{-1}B = \frac{1}{-2}\begin{pmatrix} -2 & -4 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} -11 \\ 7 \end{pmatrix}$ | (M1) | |
| $X = \begin{pmatrix} 3 \\ -5 \end{pmatrix}$ | (A1) | Must be in matrix form |
| | [4] | |

## Part b(i)):

| Working | Mark | Guidance |
|---------|------|----------|
| If reflection in $y = -2x$, then $\tan\theta = -2$, $\therefore \theta = \tan^{-1}(-2)$ | B1 | si |
| Reflection matrix: $\begin{pmatrix} -\frac{3}{5} & -\frac{4}{5} \\ -\frac{4}{5} & \frac{3}{5} \end{pmatrix}$ | B2 | B1 for 1 error (possibly repeated); If B2 then -1 for PA |
| | [3] | |

## Part b(ii)):

**Method 1:**

| Working | Mark | Guidance |
|---------|------|----------|
| $EF = \begin{pmatrix} -\frac{3}{5} & -\frac{4}{5} \\ -\frac{4}{5} & \frac{3}{5} \end{pmatrix}\begin{pmatrix} 2 & 3 \\ 7 & 1 \end{pmatrix}$ | M1 | FT their $T$; for attempt to multiply at least 1 point matrix |
| $EF = \begin{pmatrix} -\frac{34}{5} & -\frac{13}{5} \\ \frac{13}{5} & -\frac{9}{5} \end{pmatrix}$ | A1, A1 | Left column; Right column (may be seen as separate matrices) |
| Midpoint: $\left(-\frac{47}{10}, \frac{2}{5}\right)$ | B1 | oe, FT their $E$ and $F$ |

**Method 2:**

| Working | Mark | Guidance |
|---------|------|----------|
| Midpoint of $CD = \left(\frac{2+3}{2}, \frac{7+1}{2}\right) = \left(\frac{5}{2}, 4\right)$ | (B1) | FT their $T$ |
| $\begin{pmatrix} -\frac{3}{5} & -\frac{4}{5} \\ -\frac{4}{5} & \frac{3}{5} \end{pmatrix}\begin{pmatrix} \frac{5}{2} \\ 4 \end{pmatrix}$ | (M1) | FT their midpoint |
| $= \begin{pmatrix} -\frac{47}{10} \\ \frac{2}{5} \end{pmatrix}$ | (A1) | |
| Midpoint of $EF$: $\left(-\frac{47}{10}, \frac{2}{5}\right)$ | (A1) | oe |
| | [4] | |
| | **[11]** | |

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2. (a) The matrices $\mathbf { A }$ and $\mathbf { B }$ are defined by

$$\mathbf { A } = \left( \begin{array} { c c } 
3 & 4 \\
- 1 & - 2
\end{array} \right) , \quad \mathbf { B } = \binom { - 11 } { 7 }$$

Given that $\mathbf { A X } = \mathbf { B }$, find the matrix $\mathbf { X }$.\\
(b) (i) Find the $2 \times 2$ matrix, $\mathbf { T }$, which represents a reflection in the line $y = - 2 x$.\\
(ii) The images of the points $C ( 2,7 )$ and $D ( 3,1 )$, under $\mathbf { T }$, are $E$ and $F$ respectively. Find the coordinates of the midpoint of $E F$.\\

\hfill \mbox{\textit{WJEC Further Unit 1 2022 Q2 [11]}}

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