Inverse transformations
Learn how inverse transformations undo a linear transformation in Year 11 Specialist Mathematics for Queensland (QCAA). 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.
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 (QCAA, Queensland). 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.