The Inverse of a Linear Transformation
Undoing a transformation is exactly matrix inversion — but singular transformations destroy information, and once it's gone, no inverse can bring it back.
Inverse transformations
If a transformation is represented by matrix $M$, the transformation that undoes it is represented by $M^{-1}$, provided $\det(M)\neq0$. If $\det(M)=0$, the transformation is singular: it collapses the plane onto a line (or a point), so information is lost and no inverse transformation exists.
A singular $2\times2$ matrix maps every point in the plane onto a single line through the origin. Two different starting points can map to the same image point, so there is no way to reverse the transformation uniquely — this is why $M^{-1}$ cannot exist when $\det(M)=0$.
For the standard transformations, the inverse can usually be described geometrically without any calculation:
- Enlargement, scale factor $k$: the inverse is an enlargement, scale factor $\dfrac1k$ (undoing "grow by $k$" means "shrink by a factor of $k$", i.e. grow by $\tfrac1k$).
- Rotation by angle $\theta$ anticlockwise: the inverse is a rotation by $\theta$ clockwise, equivalently a rotation by $-\theta$ (or $360°-\theta$) anticlockwise.
- Reflection in any line: reflections are self-inverse — reflecting twice in the same line returns every point to where it started, so $M^{-1}=M$ for any reflection matrix $M$.
- Stretch, scale factor $a$ parallel to one axis: the inverse is a stretch, scale factor $\dfrac1a$, parallel to the same axis.
$\det(M)=0(0)-1(1)=-1$, so $M^{-1}=\dfrac{1}{-1}\begin{pmatrix}0&-1\\-1&0\end{pmatrix}=\begin{pmatrix}0&1\\1&0\end{pmatrix}=M$.
This confirms the general geometric rule: reflecting a point in $y=x$ and then reflecting the result in $y=x$ again returns it to its starting position.
$\det(M)=2(1)-1(1)=1$, so $M^{-1}=\begin{pmatrix}1&-1\\-1&2\end{pmatrix}$.
The point $(2,1)$ maps to $(5,3)$ under $M$ — check: $M\begin{pmatrix}2\\1\end{pmatrix}=\begin{pmatrix}4+1\\2+1\end{pmatrix}=\begin{pmatrix}5\\3\end{pmatrix}$ ✓.
$\det(M)=2(2)-4(1)=0$, so $M$ is singular and has no inverse.
The columns $(2,1)$ and $(4,2)$ are both scalar multiples of $(1,\tfrac12)$ — they are parallel. The unit square is squashed onto the line through the origin with direction $(2,1)$; the transformation has zero area scale factor.
Inverses of composite transformations
$$(AB)^{-1}=B^{-1}A^{-1}$$ The order reverses, mirroring the fact that to undo "put on socks, then shoes" you must first take off the shoes, then the socks.
By Theorem 29.2, $(M_2M_1)^{-1}=M_1^{-1}M_2^{-1}$.
$M_1=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$, so $M_1^{-1}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ (rotation by $-90°$). $M_2^{-1}=\begin{pmatrix}\tfrac12&0\\0&\tfrac12\end{pmatrix}$.
Combined transformation is $CBA$; its inverse is $(CBA)^{-1}=A^{-1}B^{-1}C^{-1}$ (reverse order, each individually inverted).
$A^{-1}=\begin{pmatrix}\tfrac13&0\\0&1\end{pmatrix}$, $B^{-1}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$, $C^{-1}=\begin{pmatrix}1&0\\0&\tfrac12\end{pmatrix}$.
Applying to $(6,4)$: first $C^{-1}\begin{pmatrix}6\\4\end{pmatrix}=\begin{pmatrix}6\\2\end{pmatrix}$; then $B^{-1}\begin{pmatrix}6\\2\end{pmatrix}=\begin{pmatrix}2\\-6\end{pmatrix}$; then $A^{-1}\begin{pmatrix}2\\-6\end{pmatrix}=\begin{pmatrix}\tfrac23\\-6\end{pmatrix}$.
The original point was $\left(\tfrac23,-6\right)$.
The inverse of a product reverses order: $(AB)^{-1}=B^{-1}A^{-1}$, not $A^{-1}B^{-1}$ — this remains true no matter how many transformations are composed.
Exam-style questions
Question 1 5 marks
Find the value of $k$ for which $M=\begin{pmatrix}3&k\\2&4\end{pmatrix}$ is singular, and explain what this means geometrically.
$\det(M)=3(4)-k(2)=12-2k$ (M1)
Singular when $12-2k=0 \Rightarrow k=6$ (M1, A1)
When $k=6$, the transformation maps the whole plane onto a single line through the origin, so it has no inverse. (B1, B1)
Question 2 6 marks
$M=\begin{pmatrix}1&2\\3&5\end{pmatrix}$ transforms point $P$ to the point $(7,16)$. Find the coordinates of $P$.
$\det(M)=1(5)-2(3)=-1$ (M1, A1)
$M^{-1}=\dfrac{1}{-1}\begin{pmatrix}5&-2\\-3&1\end{pmatrix}=\begin{pmatrix}-5&2\\3&-1\end{pmatrix}$ (M1)
$P=M^{-1}\begin{pmatrix}7\\16\end{pmatrix}=\begin{pmatrix}-35+32\\21-16\end{pmatrix}=\begin{pmatrix}-3\\5\end{pmatrix}$ (M1, A1)
$P=(-3,5)$ (A1)
Question 3 6 marks
Given $A=\begin{pmatrix}2&0\\0&3\end{pmatrix}$ and $B=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$, find $(AB)^{-1}$ using the reversal rule, and verify your answer by computing $AB$ and inverting it directly.
$A^{-1}=\begin{pmatrix}\tfrac12&0\\0&\tfrac13\end{pmatrix}$, $B^{-1}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ (M1, A1)
$(AB)^{-1}=B^{-1}A^{-1}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}\tfrac12&0\\0&\tfrac13\end{pmatrix}=\begin{pmatrix}0&\tfrac13\\-\tfrac12&0\end{pmatrix}$ (M1, A1)
Direct check: $AB=\begin{pmatrix}2&0\\0&3\end{pmatrix}\begin{pmatrix}0&-1\\1&0\end{pmatrix}=\begin{pmatrix}0&-2\\3&0\end{pmatrix}$, $\det=6$, $(AB)^{-1}=\dfrac16\begin{pmatrix}0&2\\-3&0\end{pmatrix}=\begin{pmatrix}0&\tfrac13\\-\tfrac12&0\end{pmatrix}$ ✓ (M1, A1)