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

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

A scalar multiplies every entry of a matrix.
For matrix multiplication, combine each row of the first matrix with each column of the second using multiply-then-add.
The product $AB$ is defined when the number of columns of $A$ equals the number of rows of $B$ ; here calculations use only $2\times2$ and $2\times1$ matrices.
Matrix multiplication is order-sensitive: in general $AB\ne BA$ .

1. Work out $3\begin{pmatrix}2&-1\\0&4\end{pmatrix}$ .[1 mark]
Answer
- $\begin{pmatrix}6&-3\\0&12\end{pmatrix}$
Method: Multiply every entry by $3$ : $3(2)=6$ , $3(-1)=-3$ , $3(0)=0$ and $3(4)=12$ .

1. Work out $\begin{pmatrix}2&-1\\3&4\end{pmatrix}\begin{pmatrix}5\\-2\end{pmatrix}$ .[2 marks]
Answer
- $\begin{pmatrix}12\\7\end{pmatrix}$
Method: The top entry is $2(5)+(-1)(-2)=12$ . The bottom entry is $3(5)+4(-2)=7$ . Therefore the product is $\begin{pmatrix}12\\7\end{pmatrix}$ .

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]
Answer
- $p=3$ , $q=3$
Method: 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 $-1+4q=11$ .

The $2\times2$ identity matrix is $I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ .
It leaves a compatible matrix unchanged: $AI=IA=A$ .
For a column vector, multiplying by $I$ leaves the represented point unchanged.
Do not confuse the identity matrix with the zero matrix; its diagonal entries are $1$ .

1. Write down the $2\times2$ identity matrix $I$ .[1 mark]
Answer
- $I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$
Method: Place $1$ s on the main diagonal and $0$ s elsewhere: $I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ .

1. Let $A=\begin{pmatrix}2&-3\\5&1\end{pmatrix}$ . Work out both $AI$ and $IA$ .[2 marks]
Answer
- $AI=IA=\begin{pmatrix}2&-3\\5&1\end{pmatrix}$
Method: 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.

1. Let $P=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ . Show that $P^2=I$ and hence work out $P^{17}$ .[3 marks]
Answer
- $P^2=I$
- $P^{17}=\begin{pmatrix}0&1\\1&0\end{pmatrix}$
Method: $P^2=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I$ . Therefore $P^{17}=(P^2)^8P=I^8P=P$ .

The columns of a transformation matrix are the images of $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ respectively.
Standard matrices should be linked to rotations about the origin, reflections in $x=0$ , $y=0$ , $y=x$ or $y=-x$ , and enlargements centred at the origin.
To transform a point $(x,y)$ , multiply the matrix by the column vector $\begin{pmatrix}x\\y\end{pmatrix}$ .
When describing a transformation, give its type and all defining details: angle and direction, mirror line, or scale factor and centre.

1. Write down the matrix for a rotation of $90^\circ$ anticlockwise about the origin.[2 marks]
Answer
- $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$
Method: 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 $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ .

1. Describe fully the transformation represented by $\begin{pmatrix}-1&0\\0&1\end{pmatrix}$ and work out the image of $(4,-3)$ .[3 marks]
Answer
- Reflection in the line $x=0$
- Image: $(-4,-3)$
Method: 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 $\begin{pmatrix}-4\\-3\end{pmatrix}$ .

If matrix $A$ acts first and matrix $B$ acts second, the combined matrix is $BA$ .
Multiply transformation matrices using the usual row-by-column rule; order matters because generally $AB\ne BA$ .
A combined matrix may simplify to a familiar single transformation, which should be described fully when requested.
A common error is to write matrices in the order transformations are stated rather than in reverse action order.

1. A reflection in the $x$ -axis is followed by a reflection in the $y$ -axis. Work out the combined matrix.[2 marks]
Answer
- $\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$
Method: 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 $R_yR_x=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ .

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]
Answer
- $\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$
- Reflection in the line $y=-x$
Method: 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 $y=-x$ .