Matrix representations of standard transformationsEdexcel International A Level Further Maths: Revision notes

A $2\times2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ transforms a point $(x,y)$, written as the column vector $\begin{pmatrix} x \\ y \end{pmatrix}$, into its image: $\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} ax+by \\ cx+dy \end{pmatrix}.$ Because every point is $x(1,0)+y(0,1)$, the matrix is fixed by what it does to the two unit vectors: the first column is the image of $(1,0)$ and the second column is the image of $(0,1)$. To find the matrix of a transformation, work out where $(1,0)$ and $(0,1)$ go and write the results as columns. Every transformation here is linear, so the origin stays fixed.

Use the images of $(1,0)$ and $(0,1)$ to build each matrix:

• Reflection in the $x$-axis: $(1,0)\to(1,0)$ and $(0,1)\to(0,-1)$, so $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$.
• Reflection in the $y$-axis: $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$.
• Reflection in $y=x$: the coordinates swap, $(x,y)\to(y,x)$, so $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$.
• Reflection in $y=-x$: $(x,y)\to(-y,-x)$, so $\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$. Example: the image of $(3,-2)$ in $y=x$ is $(-2,3)$; in $y=-x$ it is $(2,-3)$. Points on the mirror line are invariant.

A rotation through angle $\theta$ anticlockwise about $O$ sends $(1,0)\to(\cos\theta,\sin\theta)$ and $(0,1)\to(-\sin\theta,\cos\theta)$: $\mathbf{R}_\theta=\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}.$ A clockwise rotation uses a negative angle. Special cases: $90^\circ$ anticlockwise is $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, $180^\circ$ is $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$, and $90^\circ$ clockwise is $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$. For $60^\circ$ the exact entries are $\cos60^\circ=\frac12$ and $\sin60^\circ=\frac{\sqrt3}{2}$. To identify a rotation from its matrix, read $\cos\theta$ from the leading diagonal and $\sin\theta$ from the bottom-left entry.

A stretch parallel to the $x$-axis with scale factor $k$ multiplies every $x$-coordinate by $k$ and leaves $y$ unchanged, so its matrix is $\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}$. A stretch parallel to the $y$-axis with scale factor $k$ is $\begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}$. In a stretch the line of invariant points is the axis perpendicular to the stretch direction (the $y$-axis for a stretch parallel to the $x$-axis). An enlargement with centre $O$ and scale factor $k$ ($k\ne0$, real) has matrix $\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$. If $k<0$ the image is on the opposite side of $O$, and $k=-1$ is the same as a half-turn.

To transform a shape, multiply the matrix by the position vector of each vertex. To identify a transformation from a matrix, decide which standard form it is: diagonal with entries $k,1$ or $1,k$ is a stretch; diagonal with equal entries is an enlargement; $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ is a rotation; and the four matrices listed above are the reflections. Worked example: a stretch parallel to the $x$-axis maps $(4,-3)$ to $(10,-3)$, so the scale factor is $\frac{10}{4}=\frac52$ and the matrix is $\begin{pmatrix} \frac52 & 0 \\ 0 & 1 \end{pmatrix}$. To find the image of a line, take a general point on it, such as $(t,2t+3)$, transform it and eliminate $t$.