Combined and inverse transformationsEdexcel International A Level Further Maths: Revision notes

If transformation $\mathbf{A}$ is applied first and then $\mathbf{B}$, a point $\mathbf{x}$ goes to $\mathbf{A}\mathbf{x}$ and then to $\mathbf{B}(\mathbf{A}\mathbf{x})$. The single matrix for the combination is therefore the product $\mathbf{BA}$: the first transformation is the matrix nearest the vector, on the right. Matrix multiplication is not commutative, so $\mathbf{BA}\ne\mathbf{AB}$ in general; the order of the transformations matters. Example: $\mathbf{P}=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ (rotation $90^\circ$ anticlockwise) followed by $\mathbf{Q}=\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$ (stretch, factor $2$, parallel to $x$) is $\mathbf{QP}=\begin{pmatrix} 0 & -2 \\ 1 & 0 \end{pmatrix}$, but $\mathbf{Q}$ followed by $\mathbf{P}$ is $\mathbf{PQ}=\begin{pmatrix} 0 & -1 \\ 2 & 0 \end{pmatrix}$. A combination of three transformations is a product of three matrices, read right to left.

Multiply the matrices first, then apply the single result to points: this is quicker than transforming each point twice. To find the image of a line, take a general point $(t,mt+c)$, apply the combined matrix and eliminate $t$. To identify a combination, work out its matrix and compare it with the standard forms. For instance reflection in $y=x$ followed by the stretch $\begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}$ gives $\begin{pmatrix} 0 & 1 \\ 3 & 0 \end{pmatrix}$ (not itself a standard single transformation).

The inverse of a transformation $\mathbf{M}$ undoes it, so $\mathbf{M}^{-1}\mathbf{M}=\mathbf{M}\mathbf{M}^{-1}=\mathbf{I}$. For $\mathbf{M}=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ with $\det\mathbf{M}=ad-bc$: $\mathbf{M}^{-1}=\frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.$ Swap the leading-diagonal entries, change the signs of the others and divide by the determinant. The inverse exists only when $\det\mathbf{M}\ne0$; if $\det\mathbf{M}=0$ the matrix is singular and no inverse transformation exists. The inverse of a combination reverses the order: $(\mathbf{BA})^{-1}=\mathbf{A}^{-1}\mathbf{B}^{-1}$. Geometrically, the inverse of a rotation through $\theta$ is a rotation through $-\theta$, a reflection is its own inverse, a stretch of factor $k$ is undone by factor $\frac1k$, and an enlargement of factor $k$ by factor $\frac1k$. To find the point mapped onto a given image, multiply the image by $\mathbf{M}^{-1}$.

The unit square with vertices $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$ is mapped to a parallelogram of area $|ad-bc|$. So $|\det\mathbf{M}|$ is the area scale factor of the transformation: every region has its area multiplied by $|\det\mathbf{M}|$. A negative determinant means the transformation reverses orientation (a reflection is involved), such as the reflections above with $\det=-1$. When $\det\mathbf{M}=0$ the plane collapses onto a line or a point, which is why no inverse exists. For a combination, $\det(\mathbf{BA})=\det\mathbf{B}\times\det\mathbf{A}$, so area scale factors multiply. Example: $\mathbf{M}=\begin{pmatrix} 4 & 3 \\ 2 & 2 \end{pmatrix}$ has $\det\mathbf{M}=8-6=2$, so a triangle of area $7$ is mapped to one of area $14$.