Matrix Transformations (all calculations restricted to 2x2 or 2x1 matrices)

M Matrix Transformations (all calculations restricted to 2x2 or 2x1 matrices) — revision question pack

4 specification points · notes, questions, answers and worked methods

Checked against AQA 8365 section M. Review basis: the qualification registry sourced from the AQA Level 2 Certificate in Further Mathematics (8365) specification; registry verification recorded 11 July 2026.

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M1 · Multiplication of matrices (2x2 by 2x2 or by 2x1), and multiplication by a scalar

Explanation

Multiplying a matrix by a scalar multiplies every entry by that number. Matrix multiplication uses row-by-column products: each entry in the result is found by multiplying corresponding entries from one row of the first matrix and one column of the second, then adding.
A product $AB$ is defined only when the number of columns of $A$ equals the number of rows of $B$ .
In this specification, calculations are restricted to $2\times2$ matrices and $2\times1$ column matrices.
The order matters because matrix multiplication is generally not commutative: $AB\ne BA$ .
Examiners expect the working for individual entries to be visible, especially where negative numbers occur.

Worked example

Given $A=\begin{pmatrix}2&-1\\3&4\end{pmatrix}$ and $B=\begin{pmatrix}1&5\\-2&0\end{pmatrix}$ , work out $AB$ .

1.First row: $2(1)+(-1)(-2)=4$ and $2(5)+(-1)(0)=10$ .
2.Second row: $3(1)+4(-2)=-5$ and $3(5)+4(0)=15$ .
3.Place the four row-by-column results in their corresponding positions.

Answer: $AB=\begin{pmatrix}4&10\\-5&15\end{pmatrix}$ .

Common mistakes

Don't multiply entries in the same positions instead of forming row-by-column products.
Don't lose the negative sign in $(-1)(-2)$ or $4(-2)$ .
Don't assume $AB=BA$ and reverse the order without recalculating.

Exam tip

For a matrix product, show at least one complete row-by-column calculation before writing the resulting matrix.

Tier 1 · Easy

1
Work out $3\begin{pmatrix}2&-1\\0&4\end{pmatrix}$ .
[1 mark]
2
Let $A=\begin{pmatrix}1&3\\-2&4\end{pmatrix}$ and $B=\begin{pmatrix}5&-2\\0&4\end{pmatrix}$ . Work out the top-right entry of $AB$ .
[1 mark]

Tier 2 · Standard

1
Work out $\begin{pmatrix}2&-1\\3&4\end{pmatrix}\begin{pmatrix}5\\-2\end{pmatrix}$ .
[2 marks]
2
Given $A=\begin{pmatrix}1&2\\-1&3\end{pmatrix}$ and $B=\begin{pmatrix}2&0\\4&-1\end{pmatrix}$ , work out $BA$ .
[2 marks]
3
Let $A=\begin{pmatrix}-1&2\\3&1\end{pmatrix}$ and $\mathbf{u}=\begin{pmatrix}4\\-2\end{pmatrix}$ . Work out $-2A\mathbf{u}$ .
[3 marks]

Tier 3 · Hard

1
Given $\begin{pmatrix}p&2\\1&q\end{pmatrix}\begin{pmatrix}3&-1\\2&4\end{pmatrix}=\begin{pmatrix}13&5\\9&11\end{pmatrix}$ , work out $p$ and $q$ .
[4 marks]
2
Let $A=\begin{pmatrix}1&-2\\3&0\end{pmatrix}$ and $B=\begin{pmatrix}2&1\\-1&2\end{pmatrix}$ . Work out $AB$ . Hence work out the vector $\mathbf{v}=\begin{pmatrix}x\\y\end{pmatrix}$ such that $AB\mathbf{v}=\begin{pmatrix}19\\21\end{pmatrix}$ .
[4 marks]
3
$A=\begin{pmatrix}1&p\\0&2\end{pmatrix}$ and $B=\begin{pmatrix}3&1\\0&4\end{pmatrix}$ . Given that $AB=BA$ , work out $p$ .
[4 marks]
4
Let $A=\begin{pmatrix}2&-1\\3&0\end{pmatrix}$ and $\mathbf{v}=\begin{pmatrix}p\\2\end{pmatrix}$ , where $p$ is a constant. Given that $3A\mathbf{v}=\begin{pmatrix}6\\18\end{pmatrix}$ , work out $p$ .
[3 marks]
5
Let $A=\begin{pmatrix}1&-1\\2&0\end{pmatrix}$ , $B=\begin{pmatrix}0&2\\3&1\end{pmatrix}$ and $C=\begin{pmatrix}1&0\\-1&2\end{pmatrix}$ . Work out $-2ABC$ .
[3 marks]

M2 · The identity matrix I (2x2 only)

Explanation

The $2\times2$ identity matrix is $I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ . It plays the same role in matrix multiplication that $1$ plays in ordinary multiplication: for every compatible matrix $A$ , $AI=IA=A$ .
Multiplying a $2\times1$ column vector by $I$ therefore leaves the represented point unchanged. The positions of the entries matter: the $1$ s lie on the main diagonal and the other entries are $0$ .
The identity matrix is not the zero matrix.
It can also appear when a transformation and its inverse are combined, or when a matrix power cycles back to the starting position.
Examiners may require the identity to be recognised, written down or used in a chain of powers.

Worked example

Let $P=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ . Show that $P^2=I$ and hence work out $P^{12}$ .

1. $P^2=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I$ .
2.Write $P^{12}=(P^2)^6$ .
3.Therefore $P^{12}=I^6=I$ .

Answer: $P^{12}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ .

Common mistakes

Don't write $I=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ by putting the $1$ s on the wrong diagonal.
Don't treat $I$ as the zero matrix and conclude that $AI$ is zero.
Don't expand $P^{12}$ as twelve separate multiplications instead of using $P^2=I$ .

Exam tip

When a question says ‘hence’, use the established identity relation explicitly rather than restarting the calculation.

Tier 1 · Easy

1
Write down the $2\times2$ identity matrix $I$ .
[1 mark]
2
Work out $I\begin{pmatrix}7\\-4\end{pmatrix}$ , where $I$ is the $2\times2$ identity matrix.
[1 mark]

Tier 2 · Standard

1
Let $A=\begin{pmatrix}2&-3\\5&1\end{pmatrix}$ . Work out both $AI$ and $IA$ .
[2 marks]
2
The matrix $C=\begin{pmatrix}a&-2\\5&b\end{pmatrix}$ satisfies $CI=\begin{pmatrix}4&-2\\5&7\end{pmatrix}$ . Work out $a$ and $b$ .
[2 marks]
3
The matrix $X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ satisfies $\begin{pmatrix}1&2\\0&1\end{pmatrix}X=\begin{pmatrix}1&2\\0&1\end{pmatrix}$ . Work out $X$ .
[3 marks]

Tier 3 · Hard

1
Let $P=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ . Show that $P^2=I$ and hence work out $P^{17}$ .
[3 marks]
2
Let $Q=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ . Show that $Q^2=-I$ and hence work out $Q^{202}$ .
[4 marks]
3
Let $A=\begin{pmatrix}p&2\\3&-p\end{pmatrix}$ . Show that $A^2=(p^2+6)I$ . Given that $A^2=10I$ , work out the possible values of $p$ .
[4 marks]
4
Let $A=\begin{pmatrix}1&p\\0&1\end{pmatrix}$ and $B=\begin{pmatrix}1&q\\0&1\end{pmatrix}$ . Given that $AB=I$ and $p=2q+3$ , work out $p$ and $q$ .
[4 marks]
5
Let $P=\begin{pmatrix}0&2\\\frac{1}{2}&0\end{pmatrix}$ and $X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ . Given that $PXP=I$ , work out $X$ .
[4 marks]

M3 · Transformations of the unit square in the x-y plane, represented by a 2x2 matrix (rotations of 90/180/270 about the origin, reflections in x=0, y=0, y=x, y=-x, enlargements centred on the origin)

Explanation

A $2\times2$ matrix represents a transformation of the unit square in the coordinate plane.
Its first column is the image of $\begin{pmatrix}1\\0\end{pmatrix}$ and its second column is the image of $\begin{pmatrix}0\\1\end{pmatrix}$ .
Multiplying the matrix by $\begin{pmatrix}x\\y\end{pmatrix}$ gives the image of $(x,y)$ .
Required transformations are rotations of $90^\circ$ , $180^\circ$ or $270^\circ$ about the origin; reflections in $x=0$ , $y=0$ , $y=x$ or $y=-x$ ; and enlargements centred on the origin.
A full description gives the angle and centre of a rotation, adding direction for $90^\circ$ or $270^\circ$ ; the mirror line of a reflection; or the scale factor and centre of an enlargement.

The unit square and its image after a 90° anticlockwise rotation about the origin.

Worked example

The matrix $T=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ acts on the point $(3,-2)$ . Work out the image and describe geometrically the single transformation represented by $T$ .

1. $T\begin{pmatrix}3\\-2\end{pmatrix}=\begin{pmatrix}0(3)+(-1)(-2)\\1(3)+0(-2)\end{pmatrix}=\begin{pmatrix}2\\3\end{pmatrix}$ .
2.The columns show $(1,0)\mapsto(0,1)$ and $(0,1)\mapsto(-1,0)$ .
3.These images identify a rotation of $90^\circ$ anticlockwise about the origin.

Answer: The image is $(2,3)$ ; the transformation is a rotation of $90^\circ$ anticlockwise about the origin.

Common mistakes

Don't read the rows, rather than the columns, as the images of the two unit vectors.
Don't describe a $90^\circ$ rotation without stating its direction and centre.
Don't use $(x,y)T$ instead of multiplying $T\begin{pmatrix}x\\y\end{pmatrix}$ .

Exam tip

To identify an unfamiliar matrix, map the two unit vectors and then give the transformation with all defining details.

Tier 1 · Easy

1
Write down the matrix for a rotation of $90^\circ$ anticlockwise about the origin.
[2 marks]
2
The matrix $T=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ acts on the point $(2,-5)$ . Work out the image of the point.
[2 marks]

Tier 2 · Standard

1
Describe geometrically the single transformation represented by $\begin{pmatrix}-1&0\\0&1\end{pmatrix}$ and work out the image of $(4,-3)$ .
[3 marks]
2
An enlargement centred on the origin maps $(2,-1)$ to $(6,-3)$ . Work out its scale factor and write down its matrix.
[2 marks]
3
Describe geometrically the single transformation represented by $T=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ . The image of a point $P$ under $T$ is $(-4,7)$ . Work out the coordinates of $P$ .
[3 marks]

Tier 3 · Hard

1
A transformation maps $\begin{pmatrix}1\\0\end{pmatrix}$ to $\begin{pmatrix}0\\-1\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ to $\begin{pmatrix}-1\\0\end{pmatrix}$ . Write its matrix, describe geometrically the single transformation represented by this matrix, and work out the image of $(3,-2)$ .
[4 marks]
2
A matrix $T$ maps $(1,1)$ to $(1,-1)$ and maps $(1,-1)$ to $(-1,-1)$ . Work out $T$ . Describe geometrically the single transformation represented by $T$ , and work out the image of $(-3,2)$ .
[4 marks]
3
Describe geometrically the single transformation represented by $T=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ . A point $P=(a,b)$ is unchanged by $T$ and lies on the line $x-y=6$ . Work out the coordinates of $P$ .
[3 marks]
4
A matrix $T$ maps $(1,0)$ to $(-3,0)$ and $(0,1)$ to $(0,-3)$ . Work out $T$ and $T^2$ . Describe geometrically the single transformations represented by $T$ and by $T^2$ .
[4 marks]
5
Describe geometrically the single transformation represented by $T=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ . A point $P=(x,y)$ lies on $y=2x+1$ . Its image $Q=(X,Y)$ under $T$ lies on $Y=X+7$ . Work out the coordinates of $P$ and $Q$ .
[4 marks]

M4 · Combination of transformations using matrix multiplication

Explanation

Successive matrix transformations combine through matrix multiplication. If transformation $A$ acts first and transformation $B$ acts second, a vector $\mathbf{x}$ becomes $B(A\mathbf{x})=(BA)\mathbf{x}$ , so the combined matrix is $BA$ .
The right-hand matrix therefore acts first. Order is essential because $AB$ and $BA$ are generally different transformations.
After multiplying, the combined matrix may match a familiar rotation, reflection or enlargement and should be described fully when requested.
Calculations remain within $2\times2$ and $2\times1$ matrices; $\mathbf{i}$ and $\mathbf{j}$ notation is not required.
Examiners expect the action order to be translated into the correct matrix order before row-by-column multiplication begins.

Transformation A acts first and transformation B acts second, so the combined matrix is BA.

Worked example

A reflection in the line $y=x$ is followed by a rotation of $90^\circ$ clockwise about the origin. Work out the combined matrix and describe geometrically the single transformation it represents.

1.Use $S=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ for the reflection and $R=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ for the rotation.
2.Since $S$ acts first, calculate $RS=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ .
3.The matrix leaves $x$ unchanged and changes $y$ to $-y$ .

Answer: The combined transformation is reflection in the line $y=0$ .

Common mistakes

Don't write $SR$ because the transformations are copied in the order stated, even though $S$ acts first.
Don't multiply the matrices entry by entry rather than row by column.
Don't call the combined matrix a reflection without identifying the mirror line $y=0$ .

Exam tip

Write a short action chain such as $\mathbf{x}\to A\mathbf{x}\to BA\mathbf{x}$ before forming the combined matrix.

Tier 1 · Easy

1
A reflection in the $x$ -axis is followed by a reflection in the $y$ -axis. Work out the combined matrix.
[2 marks]
2
Transformation $A$ has matrix $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ and acts first. Transformation $B$ has matrix $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and acts second. Work out the combined matrix.
[2 marks]

Tier 2 · Standard

1
Matrix $A$ rotates points $90^\circ$ anticlockwise about the origin. Matrix $B$ reflects points in the $x$ -axis. Transformation $A$ acts before transformation $B$ . Work out the combined matrix and describe its effect.
[3 marks]
2
Let $R$ be reflection in the $x$ -axis and let $Q$ be rotation by $90^\circ$ anticlockwise about the origin. Work out the combined matrix for $R$ followed by $Q$ , and the combined matrix for $Q$ followed by $R$ .
[3 marks]
3
A reflection in the line $y=-x$ is followed by a reflection in the $x$ -axis. Work out the combined matrix and the image of $(3,-1)$ .
[3 marks]

Tier 3 · Hard

1
A point is reflected in $y=x$ , enlarged by scale factor $3$ about the origin, then rotated $90^\circ$ clockwise about the origin. Work out the combined matrix, describe its geometric effect, and work out the image of $(-2,1)$ .
[4 marks]
2
A reflection in $y=x$ and a rotation of $90^\circ$ anticlockwise about the origin are applied once each, but their order is unknown. The point $(3,-2)$ maps to $(-3,-2)$ . Work out the order and the combined matrix, and describe geometrically the single transformation the combined matrix represents.
[4 marks]
3
A reflection $S$ in the $x$ -axis is applied first, followed by a transformation $U$ . The combined matrix is $C=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$ . Work out the matrix $U$ . Describe geometrically the single transformation represented by $U$ .
[4 marks]
4
A reflection in the line $y=x$ is applied, then a rotation of $90^\circ$ anticlockwise about the origin, then a reflection in the line $y=x$ again. Work out the single combined matrix. Describe geometrically the single transformation it represents.
[4 marks]
5
The transformation with matrix $C=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$ is applied to every point of the plane. A single transformation is then applied so that every point returns to its original position. Work out the matrix of the second transformation, and describe it geometrically.
[3 marks]

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

M1 · Multiplication of matrices (2x2 by 2x2 or by 2x1), and multiplication by a scalar

Tier 1 · Easy

Mark scheme for M1 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$\begin{pmatrix}6&-3\\0&12\end{pmatrix}$	1	Multiply every entry by $3$ : $3(2)=6$ , $3(-1)=-3$ , $3(0)=0$ and $3(4)=12$ .
2	$10$	1	Use the top row of $A$ and the right-hand column of $B$ : $1(-2)+3(4)=-2+12=10$ .

Tier 2 · Standard

Mark scheme for M1 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$\begin{pmatrix}12\\7\end{pmatrix}$	2	The top entry is $2(5)+(-1)(-2)=12$ . The bottom entry is $3(5)+4(-2)=7$ . Therefore the product is $\begin{pmatrix}12\\7\end{pmatrix}$ .
2	$BA=\begin{pmatrix}2&4\\5&5\end{pmatrix}$	2	Multiply each row of $B$ by each column of $A$ : $BA=\begin{pmatrix}2(1)+0(-1)&2(2)+0(3)\\4(1)+(-1)(-1)&4(2)+(-1)(3)\end{pmatrix}=\begin{pmatrix}2&4\\5&5\end{pmatrix}$ .
3	$\begin{pmatrix}16\\-20\end{pmatrix}$	3	First, $A\mathbf{u}=\begin{pmatrix}-1(4)+2(-2)\\3(4)+1(-2)\end{pmatrix}=\begin{pmatrix}-8\\10\end{pmatrix}$ . Multiplying both entries by $-2$ gives $-2A\mathbf{u}=\begin{pmatrix}16\\-20\end{pmatrix}$ .

Tier 3 · Hard

Mark scheme for M1 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$p=3$ , $q=3$	4	Multiplication gives $\begin{pmatrix}3p+4&-p+8\\3+2q&-1+4q\end{pmatrix}$ . From $3p+4=13$ , $p=3$ ; this also gives $-p+8=5$ . From $3+2q=9$ , $q=3$ ; this also gives $-1+4q=11$ .
2	$AB=\begin{pmatrix}4&-3\\6&3\end{pmatrix}$ $\mathbf{v}=\begin{pmatrix}4\\-1\end{pmatrix}$	4	First, $AB=\begin{pmatrix}1(2)+(-2)(-1)&1(1)+(-2)(2)\\3(2)+0(-1)&3(1)+0(2)\end{pmatrix}=\begin{pmatrix}4&-3\\6&3\end{pmatrix}$ . Then $AB\mathbf{v}=\begin{pmatrix}19\\21\end{pmatrix}$ gives the simultaneous equations $4x-3y=19$ and $6x+3y=21$ . Adding them, $10x=40$ , so $x=4$ ; substituting back, $6(4)+3y=21$ gives $y=-1$ . Therefore $\mathbf{v}=\begin{pmatrix}4\\-1\end{pmatrix}$ .
3	$p=1$	4	$AB=\begin{pmatrix}3&1+4p\\0&8\end{pmatrix}$ and $BA=\begin{pmatrix}3&3p+2\\0&8\end{pmatrix}$ . Equating the top-right entries gives $1+4p=3p+2$ , so $p=1$ . With this value, both products are $\begin{pmatrix}3&5\\0&8\end{pmatrix}$ .
4	$p=2$	3	$A\mathbf{v}=\begin{pmatrix}2p-2\\3p\end{pmatrix}$ , so $3A\mathbf{v}=\begin{pmatrix}6p-6\\9p\end{pmatrix}$ . The second entry gives $9p=18$ , so $p=2$ ; the first entry checks, since $6(2)-6=6$ .
5	$\begin{pmatrix}8&-4\\8&-16\end{pmatrix}$	3	Multiply from the right: $BC=\begin{pmatrix}-2&4\\2&2\end{pmatrix}$ . Then $A(BC)=\begin{pmatrix}-4&2\\-4&8\end{pmatrix}$ . Multiplying every entry by $-2$ gives $-2ABC=\begin{pmatrix}8&-4\\8&-16\end{pmatrix}$ .

M2 · The identity matrix I (2x2 only)

Tier 1 · Easy

Mark scheme for M2 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$	1	Place $1$ s on the main diagonal and $0$ s elsewhere: $I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ .
2	$\begin{pmatrix}7\\-4\end{pmatrix}$	1	The identity matrix leaves a column vector unchanged, so $I\begin{pmatrix}7\\-4\end{pmatrix}=\begin{pmatrix}7\\-4\end{pmatrix}$ .

Tier 2 · Standard

Mark scheme for M2 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$AI=IA=\begin{pmatrix}2&-3\\5&1\end{pmatrix}$	2	Multiplying on the right by $I$ preserves the columns of $A$ , so $AI=A$ . Multiplying on the left preserves its rows, so $IA=A$ . Direct row-by-column multiplication gives $\begin{pmatrix}2&-3\\5&1\end{pmatrix}$ both times.
2	$a=4$ , $b=7$	2	Multiplication by $I$ does not change a matrix, so $CI=C$ . Comparing corresponding entries with $\begin{pmatrix}4&-2\\5&7\end{pmatrix}$ gives $a=4$ and $b=7$ .
3	$X=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I$	3	The product on the left is $\begin{pmatrix}a+2c&b+2d\\c&d\end{pmatrix}$ . Comparing the bottom row gives $c=0$ and $d=1$ . The top row then gives $a=1$ and $b=0$ , so $X=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I$ .

Tier 3 · Hard

Mark scheme for M2 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$P^2=I$ $P^{17}=\begin{pmatrix}0&1\\1&0\end{pmatrix}$	3	$P^2=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I$ . Therefore $P^{17}=(P^2)^8P=I^8P=P$ .
2	$Q^2=-I=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ $Q^{202}=-I=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$	4	$Q^2=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}0&-1\\1&0\end{pmatrix}=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}=-I$ . Hence $Q^4=(-I)^2=I$ . Since $202=4(50)+2$ , $Q^{202}=(Q^4)^{50}Q^2=I^{50}(-I)=-I$ .
3	$A^2=\begin{pmatrix}p^2+6&0\\0&p^2+6\end{pmatrix}=(p^2+6)I$ $p=-2$ or $p=2$	4	Multiplication gives $A^2=\begin{pmatrix}p^2+6&2p-2p\\3p-3p&p^2+6\end{pmatrix}=\begin{pmatrix}p^2+6&0\\0&p^2+6\end{pmatrix}=(p^2+6)I$ . Therefore $p^2+6=10$ , so $p^2=4$ and $p=-2$ or $p=2$ .
4	$p=1$ , $q=-1$	4	$AB=\begin{pmatrix}1&p+q\\0&1\end{pmatrix}$ . Since $AB=I$ , the top-right entry gives $p+q=0$ . Substituting $p=2q+3$ gives $3q+3=0$ , so $q=-1$ and $p=1$ .
5	$X=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I$	4	First, $PX=\begin{pmatrix}2c&2d\\\frac{a}{2}&\frac{b}{2}\end{pmatrix}$ . Hence $PXP=\begin{pmatrix}d&4c\\\frac{b}{4}&a\end{pmatrix}$ . Equating this to $I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ gives $a=d=1$ and $b=c=0$ , so $X=I$ .

M3 · Transformations of the unit square in the x-y plane, represented by a 2x2 matrix (rotations of 90/180/270 about the origin, reflections in x=0, y=0, y=x, y=-x, enlargements centred on the origin)

Tier 1 · Easy

Mark scheme for M3 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$\begin{pmatrix}0&-1\\1&0\end{pmatrix}$	2	The vector $\begin{pmatrix}1\\0\end{pmatrix}$ maps to $\begin{pmatrix}0\\1\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ maps to $\begin{pmatrix}-1\\0\end{pmatrix}$ . These images form the columns of $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ .
2	$(-5,2)$	2	$T\begin{pmatrix}2\\-5\end{pmatrix}=\begin{pmatrix}0(2)+1(-5)\\1(2)+0(-5)\end{pmatrix}=\begin{pmatrix}-5\\2\end{pmatrix}$ , so the image is $(-5,2)$ .

Tier 2 · Standard

Mark scheme for M3 Tier 2 · Standard
Q	Answer	Mark	Comments
1	Reflection in the line $x=0$ Image: $(-4,-3)$	3	The matrix changes $x$ to $-x$ and leaves $y$ unchanged, so it is reflection in the $y$ -axis, whose equation is $x=0$ . Multiplying by $\begin{pmatrix}4\\-3\end{pmatrix}$ gives $\begin{pmatrix}-4\\-3\end{pmatrix}$ .
2	Scale factor $3$ $\begin{pmatrix}3&0\\0&3\end{pmatrix}$	2	Both coordinates have been multiplied by $3$ , so the scale factor is $3$ . An enlargement of scale factor $3$ centred on the origin is represented by $\begin{pmatrix}3&0\\0&3\end{pmatrix}$ .
3	Rotation of $180^\circ$ about the origin (accept enlargement with scale factor $-1$ centred on the origin) $P=(4,-7)$	3	The matrix sends $(x,y)$ to $(-x,-y)$ , so it represents a rotation of $180^\circ$ about the origin. If $P=(x,y)$ , then $(-x,-y)=(-4,7)$ . Hence $x=4$ and $y=-7$ , so $P=(4,-7)$ .

Tier 3 · Hard

Mark scheme for M3 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$ Reflection in the line $y=-x$ Image: $(2,-3)$	4	The two given image vectors are the columns, so the matrix is $\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$ . It sends $(x,y)$ to $(-y,-x)$ , which is reflection in $y=-x$ . Multiplying it by $\begin{pmatrix}3\\-2\end{pmatrix}$ gives $\begin{pmatrix}2\\-3\end{pmatrix}$ .
2	$T=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ Rotation of $90^\circ$ clockwise about the origin (accept rotation of $270^\circ$ anticlockwise about the origin) Image: $(2,3)$	4	Write $T=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ . The two mappings give $a+b=1$ , $a-b=-1$ , $c+d=-1$ and $c-d=-1$ . Therefore $a=0$ , $b=1$ , $c=-1$ and $d=0$ , so $T=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ . This is a rotation of $90^\circ$ clockwise about the origin, and $T\begin{pmatrix}-3\\2\end{pmatrix}=\begin{pmatrix}2\\3\end{pmatrix}$ .
3	Reflection in the line $y=0$ (accept reflection in the $x$ -axis) $P=(6,0)$	3	The matrix maps $(x,y)$ to $(x,-y)$ , so it represents reflection in the line $y=0$ , the $x$ -axis. Since $P$ is unchanged, $(a,b)=(a,-b)$ and hence $b=0$ . The line condition gives $a-0=6$ , so $a=6$ . Therefore $P=(6,0)$ .
4	$T=\begin{pmatrix}-3&0\\0&-3\end{pmatrix}$ An enlargement with scale factor $-3$ centred on the origin (accept an enlargement scale factor $3$ centred on the origin combined with a rotation of $180^\circ$ — the single-transformation answer is the enlargement with scale factor $-3$ ) $T^2=\begin{pmatrix}9&0\\0&9\end{pmatrix}$ , an enlargement with scale factor $9$ centred on the origin	4	The images of $(1,0)$ and $(0,1)$ form the columns, so $T=\begin{pmatrix}-3&0\\0&-3\end{pmatrix}=-3I$ . Every point is mapped to $-3$ times its position vector, an enlargement with scale factor $-3$ centred on the origin. Squaring, $T^2=(-3I)(-3I)=9I=\begin{pmatrix}9&0\\0&9\end{pmatrix}$ , which multiplies every position vector by $9$ : an enlargement with scale factor $9$ centred on the origin.
5	Rotation of $90^\circ$ anticlockwise about the origin $P=(2,5)$ , $Q=(-5,2)$	4	The matrix represents a rotation of $90^\circ$ anticlockwise about the origin, so $(X,Y)=(-y,x)$ . Since $Q$ lies on $Y=X+7$ , $x=-y+7$ and hence $x+y=7$ . Combining this with $y=2x+1$ gives $3x+1=7$ , so $x=2$ and $y=5$ . Therefore $P=(2,5)$ and $Q=(-5,2)$ .

M4 · Combination of transformations using matrix multiplication

Tier 1 · Easy

Mark scheme for M4 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$	2	The matrices are $R_x=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and $R_y=\begin{pmatrix}-1&0\\0&1\end{pmatrix}$ . Since $R_x$ acts first, calculate $R_yR_x=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ .
2	$BA=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$	2	Because $A$ acts first, the combined matrix is $BA$ . Therefore $BA=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ . Note that $AB=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ is different, so the order matters.

Tier 2 · Standard

Mark scheme for M4 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$ Reflection in the line $y=-x$	3	Let $A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ be the rotation and $B=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ the reflection. The combined matrix is $BA=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$ . This sends $(x,y)$ to $(-y,-x)$ , so it is reflection in $y=-x$ .
2	$R$ followed by $Q$ : $QR=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ $Q$ followed by $R$ : $RQ=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$	3	Use $R=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and $Q=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ . Then $QR=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ , whereas $RQ=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$ . The different products show that reversing the order changes the transformation.
3	$\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ (a rotation of $90^\circ$ anticlockwise about the origin) Image: $(1,3)$	3	The first reflection is $S=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$ and the second is $R=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ . Since $S$ acts first, the combined matrix is $RS=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ . Note $SR=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ differs, so the order matters. The combined matrix maps $\begin{pmatrix}3\\-1\end{pmatrix}$ to $\begin{pmatrix}1\\3\end{pmatrix}$ .

Tier 3 · Hard

Mark scheme for M4 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$\begin{pmatrix}3&0\\0&-3\end{pmatrix}$ Enlargement by scale factor $3$ about the origin combined with reflection in the $x$ -axis Image: $(-6,-3)$	4	Use $S=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ , $E=\begin{pmatrix}3&0\\0&3\end{pmatrix}$ and $R=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ . The action order gives $RES=\begin{pmatrix}3&0\\0&-3\end{pmatrix}$ . It sends $(x,y)$ to $(3x,-3y)$ , an enlargement by factor $3$ with reflection in the $x$ -axis. Applied to $(-2,1)$ it gives $(-6,-3)$ .
2	Reflection first, then rotation $\begin{pmatrix}-1&0\\0&1\end{pmatrix}$ Reflection in the line $x=0$ (accept reflection in the $y$ -axis)	4	Let $S=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ be the reflection and $Q=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ the rotation. Reflection followed by rotation gives $QS=\begin{pmatrix}-1&0\\0&1\end{pmatrix}$ , which sends $(3,-2)$ to $(-3,-2)$ . The reverse order gives $SQ=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ , which would send the point to $(3,2)$ . Therefore the reflection acts first, and the combined transformation is reflection in $x=0$ .
3	$U=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ Rotation of $90^\circ$ clockwise about the origin (accept rotation of $270^\circ$ anticlockwise about the origin)	4	Here $S=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and $S^2=I$ . Since $S$ acts first, $US=C$ . Multiplying both sides on the right by $S$ gives $USS=CS$ , so $U=CS$ . Therefore $U=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ , a rotation of $90^\circ$ clockwise about the origin.
4	$\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ Rotation of $90^\circ$ clockwise about the origin (accept rotation of $270^\circ$ anticlockwise about the origin)	4	Use $S=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ and $Q=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ . Applying $S$ first, then $Q$ , then $S$ gives combined matrix $SQS$ . First $QS=\begin{pmatrix}-1&0\\0&1\end{pmatrix}$ , and then $S(QS)=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ . Its columns send $(1,0)$ to $(0,-1)$ and $(0,1)$ to $(1,0)$ , a rotation of $90^\circ$ clockwise about the origin.
5	$\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$ (the same matrix $C$ ) Reflection in the line $y=-x$ — reflecting a second time in the same line returns every point	3	The second transformation, with matrix $U$ , must satisfy $UC=I$ . Now $C^2=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}\begin{pmatrix}0&-1\\-1&0\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I$ , so $U=C$ works. $C$ sends $(x,y)$ to $(-y,-x)$ , which is reflection in the line $y=-x$ ; reflecting twice in the same line returns every point to its original position.