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3.9 Matrix transformationsIB Maths: Applications and Interpretation HL: Revision notes

Section 1

Matrices as transformations

A 2×22\times2 matrix transforms a point by multiplying its position vector: (x′y′)=(abcd)(xy).\begin{pmatrix}x' \\ y'\end{pmatrix}=\begin{pmatrix}a & b \\ c & d\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}. The columns of the matrix are the images of (1,0)(1,0) and (0,1)(0,1), so you can write a matrix straight from a picture. All these transformations fix the origin.

Standard matrices:

  • Reflection in the xx-axis: (100−1)\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}; in the yy-axis: (−1001)\begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix}
  • Reflection in y=xy=x: (0110)\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}; in y=−xy=-x: (0−1−10)\begin{pmatrix}0 & -1 \\ -1 & 0\end{pmatrix}
  • Enlargement, scale factor kk, centre the origin: (k00k)\begin{pmatrix}k & 0 \\ 0 & k\end{pmatrix}
  • Horizontal stretch, factor kk: (k001)\begin{pmatrix}k & 0 \\ 0 & 1\end{pmatrix}; vertical stretch, factor kk: (100k)\begin{pmatrix}1 & 0 \\ 0 & k\end{pmatrix}
  • Rotation through θ\theta anticlockwise about the origin: (cos⁡θ−sin⁡θsin⁡θcos⁡θ)\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}, so 90∘90^\circ anticlockwise is (0−110)\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}

Example: (0−110)(32)=(−23)\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}3 \\ 2\end{pmatrix}=\begin{pmatrix}-2 \\ 3\end{pmatrix}.

Key termstransformation matrixreflectionstretchenlargementrotation
Exam tip

To find a matrix, ask where (1,0)(1,0) and (0,1)(0,1) go. Put the first image in column 1 and the second in column 2.

Common mistake

Mixing up a horizontal stretch (the xx-coordinates are multiplied) with a vertical stretch (the yy-coordinates are multiplied).

Section 2

Translations and the form Mx + t

A translation adds a vector to every point, so it cannot be written as a 2×22\times2 matrix multiplication by itself. A general transformation is (x′y′)=(abcd)(xy)+(ef).\begin{pmatrix}x' \\ y'\end{pmatrix}=\begin{pmatrix}a & b \\ c & d\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}+\begin{pmatrix}e \\ f\end{pmatrix}. Do the matrix multiplication first, then add the translation vector.

Example: (xy)→(2002)(xy)+(1−3)\begin{pmatrix}x \\ y\end{pmatrix}\to\begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}+\begin{pmatrix}1 \\ -3\end{pmatrix} sends (4,2)(4,2) to (8,4)+(1,−3)=(9,1)(8,4)+(1,-3)=(9,1).

A translation does not change lengths, angles or area.

Key termstranslation
Common mistake

Adding the translation vector before multiplying by the matrix. Multiply first, then add.

Section 3

Compositions of transformations

A composition applies one transformation after another. If A\mathbf{A} is applied first and then B\mathbf{B}, the single matrix is BA\mathbf{B}\mathbf{A} because B(Ax)=(BA)x\mathbf{B}(\mathbf{A}\mathbf{x})=(\mathbf{B}\mathbf{A})\mathbf{x}. The order matters, since matrix multiplication is not commutative.

Example: rotate 90∘90^\circ anticlockwise, M=(0−110)\mathbf{M}=\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}, then reflect in the xx-axis, R=(100−1)\mathbf{R}=\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}: RM=(0−1−10),\mathbf{R}\mathbf{M}=\begin{pmatrix}0 & -1 \\ -1 & 0\end{pmatrix}, which is a reflection in y=−xy=-x. Doing it the other way round gives MR=(0110)\mathbf{M}\mathbf{R}=\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}, a reflection in y=xy=x.

Key termscomposition
Common mistake

Writing the matrices in the order they are applied. The first transformation is the matrix closest to the vector.

Section 4

The determinant and area

For A=(abcd)\mathbf{A}=\begin{pmatrix}a & b \\ c & d\end{pmatrix}, det⁡A=ad−bc\det\mathbf{A}=ad-bc. The area scale factor of the transformation is ∣det⁡A∣|\det\mathbf{A}|: area of image=∣det⁡A∣×area of object.\text{area of image}=|\det\mathbf{A}|\times\text{area of object}.

  • If det⁡A<0\det\mathbf{A}<0, the image is also reflected (orientation reverses).
  • If det⁡A=0\det\mathbf{A}=0, the shape collapses to a line or a point.
  • Translations do not change area.

Example: N=(3122)\mathbf{N}=\begin{pmatrix}3 & 1 \\ 2 & 2\end{pmatrix} has det⁡N=6−2=4\det\mathbf{N}=6-2=4, so a triangle of area 88 becomes area 3232. For a composition, multiply the determinants: det⁡(BA)=det⁡B×det⁡A\det(\mathbf{B}\mathbf{A})=\det\mathbf{B}\times\det\mathbf{A}.

Key termsdeterminantarea scale factor
Common mistake

Using a negative determinant as an area. Use ∣det⁡A∣|\det\mathbf{A}|.

Section 5

Fractals by iteration

A fractal is built by applying transformations over and over. Each step uses matrix transformations of the form x→Mx+t\mathbf{x}\to\mathbf{M}\mathbf{x}+\mathbf{t}.

Example (Sierpinski triangle): start with a filled triangle of area A0A_0. Apply three transformations, each an enlargement of scale factor 0.50.5 followed by a different translation, and keep all three images. Each image has area 0.25A00.25A_0 since det⁡=0.25\det=0.25. After nn iterations there are 3n3^n small triangles and the shaded area is An=(34)nA0,A_n=\left(\tfrac34\right)^nA_0, a geometric sequence that tends to 0 as nn increases. Sums of lengths or areas of repeated, scaled copies are infinite geometric series, with S∞=a1−rS_\infty=\frac{a}{1-r} for ∣r∣<1|r|<1.

Repeatedly applying a matrix to a vector, xn+1=Axn\mathbf{x}_{n+1}=\mathbf{A}\mathbf{x}_n, links to Markov chains, where A\mathbf{A} is a transition matrix.

Key termsfractaliteration
Exam tip

In a fractal question, find the scale factor of lengths (kk) and of areas (k2=∣det⁡∣k^2=|\det|). The sequence of lengths or areas is geometric.

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Exam questions on 3.9 Matrix transformations

  1. Transformation TT is represented by the matrix M=(0−110)\mathbf{M}=\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}, so that a point (x,y)(x,y) maps to the point with position vector M(xy)\mathbf{M}\begin{pmatrix}x \\ y\end{pmatrix}.
    TT is followed by a reflection in the xx-axis, which has matrix R=(100−1)\mathbf{R}=\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}. Find the single matrix that represents this combined transformation.2 marks
  2. Triangle SS has area 88 cm2^2. It is transformed by the matrix N=(3122)\mathbf{N}=\begin{pmatrix}3 & 1 \\ 2 & 2\end{pmatrix} to give triangle S′S'.
    S′S' is then enlarged by scale factor 22, centre the origin, to give triangle S′′S''. Find the area of S′′S''.2 marks
  3. A transformation UU maps a point with position vector (xy)\begin{pmatrix}x \\ y\end{pmatrix} to (2002)(xy)+(1−3)\begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}+\begin{pmatrix}1 \\ -3\end{pmatrix}. Triangle ABCABC has vertices A(1,2)A(1,2), B(4,2)B(4,2) and C(1,6)C(1,6), with lengths in cm.
    Find the coordinates of the images A′A', B′B' and C′C' of the vertices of triangle ABCABC under UU.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).