Inverse transformations
Master how inverse transformations undo a linear transformation in Year 11 VCE Specialist Mathematics. Because every transformation of the plane is a matrix, its inverse is simply the matrix inverse — the transformation that maps each image back to where it came from. It sits in the Space and measurement area of study of the VCE Mathematics Study Design (VCAA), within the Transformations of the plane topic of Unit 2.
You will learn to find inverse transformation matrices, recover the pre-image of a point, reverse rotations, reflections and enlargements, and recognise singular transformations that collapse the plane and have no inverse — the groundwork for solving transformation problems later in the course.
Theory
An inverse transformation \(A^{-1}\) undoes a linear transformation \(A\) — a core idea in Year 11 Specialist Mathematics. Because a transformation is a \(2\times2\) matrix, its inverse is the matrix inverse, and it lets you recover the pre-image of any point. This page shows how to find \(A^{-1}\), reverse the standard transformations, and spot the singular case where no inverse exists.
A linear transformation of the plane is represented by a \(2\times2\) matrix \(A\); applying it sends each point \((x,y)\) to its image \((x',y')\). The inverse transformation \(A^{-1}\) is the transformation that sends every image back to where it came from, so that \(AA^{-1}=A^{-1}A=I\).
The transformation inverse is the matrix inverse: \(A^{-1}=\dfrac{1}{\det(A)}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\). Applying \(A^{-1}\) to an image point recovers its pre-image (the original point).
The standard transformations each have an obvious reverse: a rotation by \(\theta\) is undone by the rotation by \(-\theta\); a reflection is its own inverse (doing it twice returns every point); and an enlargement (dilation) by factor \(k\) is undone by the enlargement by \(\dfrac{1}{k}\).
If \(\det(A)=0\) the transformation is singular: it collapses the whole plane onto a line, so different points can share the same image and there is no inverse to undo it.
A transformation matrix \(A\) and its inverse (provided \(\det(A)\ne0\)):
Recovering the pre-image \((x,y)\) of an image \((x',y')\):
The standard transformations and their inverses:
How to reverse a transformation
- Write the matrix \(A\): read the transformation as a \(2\times2\) matrix (or use the given matrix directly).
- Compute the determinant: \(\det(A)=ad-bc\). If \(\det(A)=0\) the transformation is singular and has no inverse — stop here.
- Form \(A^{-1}\): swap the main-diagonal entries, negate the off-diagonal entries, and divide by \(\det(A)\).
- Apply it: to find the pre-image of an image point, multiply \(A^{-1}\) by that point. Check with \(A\times(\text{pre-image})=\text{image}\).
First the determinant \(ad-bc\):
| \(\det(A)\) | \(=\) | \((3)(1)-(2)(1)\) |
| \(=\) | \(3-2\) | |
| \(=\) | \(1\) |
Swap the main diagonal, negate the off-diagonal, then divide by \(\det(A)=1\):
| \(A^{-1}\) | \(=\) | \(\dfrac{1}{1}\begin{pmatrix}1 & -2\\ -1 & 3\end{pmatrix}\) |
| \(=\) | \(\begin{pmatrix}1 & -2\\ -1 & 3\end{pmatrix}\) |
\(A^{-1}=\begin{pmatrix}1 & -2\\ -1 & 3\end{pmatrix}\).
Find \(A^{-1}\) first (here \(\det(A)=(2)(2)-(1)(3)=1\)):
| \(A^{-1}\) | \(=\) | \(\begin{pmatrix}2 & -1\\ -3 & 2\end{pmatrix}\) |
The pre-image is \(A^{-1}\) applied to the image \((5,3)\):
| \(\begin{pmatrix}x\\y\end{pmatrix}\) | \(=\) | \(\begin{pmatrix}2 & -1\\ -3 & 2\end{pmatrix}\begin{pmatrix}5\\3\end{pmatrix}\) |
| \(=\) | \(\begin{pmatrix}(2)(5)+(-1)(3)\\(-3)(5)+(2)(3)\end{pmatrix}\) | |
| \(=\) | \(\begin{pmatrix}10-3\\-15+6\end{pmatrix}\) | |
| \(=\) | \(\begin{pmatrix}7\\-9\end{pmatrix}\) |
The original point is \((7,-9)\).
The inverse of a rotation by \(\theta\) is the rotation by \(-\theta\); compute it from the matrix (here \(\det(A)=1\)):
| \(\det(A)\) | \(=\) | \((0)(0)-(-1)(1)\) |
| \(=\) | \(1\) | |
| \(A^{-1}\) | \(=\) | \(\dfrac{1}{1}\begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}\) |
| \(=\) | \(\begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}\) |
\(A^{-1}=\begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}\): a rotation of \(90^\circ\) clockwise.
A transformation is singular exactly when \(\det(A)=0\); write \(ad-bc=0\) and solve for \(k\):
| \(\det(A)\) | \(=\) | \((3)(4)-(k)(2)\) |
| \(12-2k\) | \(=\) | \(0\) |
| \(2k\) | \(=\) | \(12\) |
| \(k\) | \(=\) | \(6\) |
\(k=6\); then the plane collapses onto a line, so \(A^{-1}\) does not exist.
Common pitfalls
Frequently asked questions
What is an inverse transformation?
It is the transformation \(A^{-1}\) that undoes \(A\): if \(A\) maps a point to its image, \(A^{-1}\) maps that image back to the original point, and \(AA^{-1}=I\).
How do you find the inverse of a transformation matrix?
Treat it as a matrix inverse: swap the main-diagonal entries, negate the off-diagonal entries, and divide by the determinant, \(A^{-1}=\dfrac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\).
How do you find the pre-image of a point?
Multiply the inverse matrix by the image point: \(\begin{pmatrix}x\\y\end{pmatrix}=A^{-1}\begin{pmatrix}x'\\y'\end{pmatrix}\). The result is the original point that mapped to that image.
What is the inverse of a rotation or a reflection?
A rotation by \(\theta\) is undone by the rotation by \(-\theta\). A reflection is its own inverse — reflecting twice in the same line returns every point to where it started.
When does a transformation have no inverse?
When \(\det(A)=0\). The transformation is then singular: it squashes the plane onto a line, so several points share one image and the mapping cannot be undone.
How does the determinant affect area under the inverse?
A transformation multiplies area by \(|\det(A)|\), so its inverse multiplies area by \(\dfrac{1}{|\det(A)|}\). If \(A\) triples area, \(A^{-1}\) scales it by one third.