General transformations
Bring together the whole of matrices and transformations in Year 11 Specialist Mathematics for Queensland (QCAA). A 2×2 matrix carries out rotations, reflections and dilations of the plane, while a column vector adds a translation to build any general transformation.
You will learn to find a matrix from a described effect, compose transformations with matrix products, undo them with inverses, use the determinant as an area scale factor, and transform about a point or line — the capstone that unifies the topic.
Theory
General transformations tie together the whole of matrices and transformations in Year 11 Specialist Mathematics (QCAA, Queensland). A 2×2 matrix carries out a linear map \(\mathbf{x}'=A\mathbf{x}\); a column vector adds a translation. This page shows how to describe a map, build composites, find inverses, track area with the determinant, and handle maps about a point or line.
A transformation of the plane sends each point \((x,y)\) to an image \((x',y')\). A linear transformation is written \(\mathbf{x}'=A\mathbf{x}\) for a \(2\times2\) matrix \(A\); its columns are the images of \((1,0)\) and \((0,1)\), so a described effect gives the matrix directly.
The standard linear maps are dilations \((x,y)\mapsto(ax,by)\), rotations about the origin, and reflections in a line through the origin. Applying one transformation and then another is a composition: "\(A\) then \(B\)" is the matrix product \(BA\), with the second map on the left.
The inverse transformation undoes a map and has matrix \(A^{-1}\); it exists exactly when \(\det(A)\ne0\). The determinant is the area scale factor: a region of area \(S\) has image area \(|\det(A)|\,S\). If \(\det(A)=0\) the map is singular and collapses every region onto a line.
A general (affine) transformation combines a linear part with a translation: \(\mathbf{x}'=A\mathbf{x}+\mathbf{b}\). To rotate or reflect about a centre \(\mathbf{c}\) that is not the origin, shift \(\mathbf{c}\) to \(O\), transform, then shift back: \(\mathbf{x}'=A(\mathbf{x}-\mathbf{c})+\mathbf{c}\).
A linear transformation and the columns rule (images of the basis vectors):
Composition, inverse, and the effect on area:
A general (affine) map, and a rotation or reflection about a centre \(\mathbf{c}\):
How to work a general transformation problem
- Name each map's matrix: read a described effect off the images of \((1,0)\) and \((0,1)\), or quote the standard rotation, reflection or dilation matrix.
- Combine in the right order: for "\(A\) then \(B\)" multiply as \(BA\) (second map on the left); to undo a map use \(A^{-1}\).
- Handle the centre: if the map is about a point or line not through \(O\), use \(\mathbf{x}'=A(\mathbf{x}-\mathbf{c})+\mathbf{c}\), or add the translation \(\mathbf{b}\).
- Answer the question: multiply by a point for its image, or take \(|\det|\) for the area scale factor; keep exact surds and fractions exact.
Multiply the matrix by the point written as a column vector, one entry at a time:
| \(\begin{pmatrix} x' \\ y' \end{pmatrix}\) | \(=\) | \(\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}\begin{pmatrix} 3 \\ 2 \end{pmatrix}\) |
| \(=\) | \(\begin{pmatrix} (0)(3)+(-1)(2) \\ (1)(3)+(0)(2) \end{pmatrix}\) | |
| \(=\) | \(\begin{pmatrix} -2 \\ 3 \end{pmatrix}\) |
The image is \((-2,3)\).
Write each matrix; "\(A\) then \(B\)" is the product \(BA\), with the rotation on the left:
| \(A\) | \(=\) | \(\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}\) |
| \(B\) | \(=\) | \(\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}\) |
Multiply \(BA\) (rotation times reflection):
| \(BA\) | \(=\) | \(\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}\) |
| \(=\) | \(\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}\) |
The combined matrix is \(\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}\) — a reflection in \(y=x\).
The undoing map is \(A^{-1}\); start with the determinant:
| \(\det(A)\) | \(=\) | \((2)(4)-(0)(0)\) |
| \(=\) | \(8\) |
Swap the main diagonal, negate the off-diagonal, then divide by the determinant:
| \(A^{-1}\) | \(=\) | \(\dfrac{1}{8}\begin{pmatrix}4 & 0\\ 0 & 2\end{pmatrix}\) |
| \(=\) | \(\begin{pmatrix}\dfrac{1}{2} & 0\\ 0 & \dfrac{1}{4}\end{pmatrix}\) |
The inverse is \(A^{-1}=\begin{pmatrix}\dfrac{1}{2} & 0\\ 0 & \dfrac{1}{4}\end{pmatrix}\).
Shift the centre \((1,2)\) to the origin, rotate, then shift back: \(\mathbf{x}'=R(\mathbf{x}-\mathbf{c})+\mathbf{c}\).
| \(\mathbf{x}-\mathbf{c}\) | \(=\) | \((4-1,\ 2-2)=(3,0)\) |
| \(R(3,0)\) | \(=\) | \(\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}\begin{pmatrix} 3 \\ 0 \end{pmatrix}=(0,3)\) |
| \(\mathbf{x}'\) | \(=\) | \((0,3)+(1,2)\) |
| \(=\) | \((1,5)\) |
The image is \((1,5)\).
Common pitfalls
Frequently asked questions
How do you find the matrix of a linear transformation?
The columns of the matrix are the images of the basis vectors: the first column is the image of \((1,0)\) and the second column is the image of \((0,1)\). Read them off the described effect.
In what order do you multiply matrices to compose transformations?
For "\(A\) then \(B\)", multiply as \(BA\) with the second transformation on the left. Applying the maps in the other order gives \(AB\), which is usually different.
How does a transformation change the area of a shape?
It multiplies every area by \(|\det(A)|\). For example, \(\det(A)=6\) makes each area six times larger; a determinant of \(0\) collapses the shape onto a line (area \(0\)).
How do you rotate a point about a centre that is not the origin?
Use \(\mathbf{x}'=A(\mathbf{x}-\mathbf{c})+\mathbf{c}\): subtract the centre \(\mathbf{c}\), apply the rotation matrix \(A\), then add \(\mathbf{c}\) back.
Does the order of transformations matter?
Yes. Matrix multiplication is not commutative, so reflecting then rotating is generally not the same as rotating then reflecting; the two composite matrices differ.
What does a determinant of zero mean for a transformation?
The transformation is singular: it has no inverse and it flattens the whole plane onto a line, so every region has image area zero.