Vector Transformations 2

Vector Transformations and Composite Matrices

IB Mathematics HL • Vectors and Matrices • Transformations • Composite Transformations

Lesson Overview

This lesson focuses on affine transformations of the plane of the form

$T(\mathbf{x})=P\mathbf{x}+\mathbf{q}$

In IB Paper 2, students may need to determine the translation vector, find the matrix of a transformation from images of points, split a matrix into simpler transformations, and identify the centre of a combined enlargement and rotation.

In this question, Zara studies how a triangle is transformed and then rewrites the same transformation in different ways.

⭐ Key Concepts

An affine transformation has the form

$T(\mathbf{x})=P\mathbf{x}+\mathbf{q}$

The image of the origin gives the translation vector $\mathbf{q}$ .
The columns of $P$ can be found from the images of $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ .
An enlargement with scale factor $k$ , centre $O$ , is represented by

$\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$

A rotation matrix has the form

$\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \quad\text{or}\quad \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$

If a point $\begin{pmatrix}a\\b\end{pmatrix}$ is the centre of a transformation, then it is invariant, so it satisfies

$P\begin{pmatrix}a\\b\end{pmatrix}+\mathbf{q}=\begin{pmatrix}a\\b\end{pmatrix}$

📘 Clear IB-Style Explanation

When a transformation is written as

$T(\mathbf{x})=P\mathbf{x}+\mathbf{q},$

the matrix $P$ controls the linear part of the transformation, while $\mathbf{q}$ shifts every point by the same vector.

A useful IB strategy is to start with the image of the origin, since

$T(\mathbf{0})=\mathbf{q}.$

After that, use the images of the basis vectors $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ to reconstruct the matrix $P$ .

Once $P$ is known, you can often factor it into a product of simpler matrices such as an enlargement and a rotation.

📌 Worked Example — Affine Transformation of a Triangle

A transformation $T$ of the plane is represented by

$T(\mathbf{x})=P\mathbf{x}+\mathbf{q},$

where $P$ is a $2\times2$ matrix and $\mathbf{q}$ is a $2\times1$ vector.

The triangle $OAB$ has coordinates

$O(0,0),\qquad A(0,1),\qquad B(1,0).$

Under $T$ , these points are transformed to

$O'(0,1),\qquad A'\left(\frac14,1+\frac{\sqrt3}{4}\right),\qquad B'\left(\frac{\sqrt3}{4},\frac34\right)$

(a.i) By considering the image of $O$ , find $\mathbf{q}$ .

(a.ii) By considering the images of $A$ and $B$ , show that

$P= \begin{pmatrix} \frac{\sqrt3}{4} & \frac14 \\ -\frac14 & \frac{\sqrt3}{4} \end{pmatrix}.$

$P$ can be written as $RS$ , where $R$ and $S$ are matrices.

$S$ represents an enlargement with scale factor $\frac12$ , centre $O$ .

$R$ represents a rotation about $O$ .

(b) Write down the matrix $S$ .

(c.i) Use $P=RS$ to find the matrix $R$ .

(c.ii) Hence find the angle and direction of the rotation represented by $R$ .

The transformation $T$ can also be described by an enlargement of scale factor $\frac12$ , centre $(a,b)$ , followed by a rotation about the same centre.

(d.i) Write down an equation satisfied by $\begin{pmatrix}a\\b\end{pmatrix}$ .

(d.ii) Find the value of $a$ and the value of $b$ .

Solution

(a.i) Find $\mathbf{q}$

Since $T(\mathbf{0})=\mathbf{q}$ , and the origin maps to $(0,1)$ ,

$\mathbf{q}= \begin{pmatrix} 0 \\ 1 \end{pmatrix}$

📊 Marks: A1

(a.ii) Find $P$

Since

$T\begin{pmatrix}1\\0\end{pmatrix} = P\begin{pmatrix}1\\0\end{pmatrix} + \begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix} \frac{\sqrt3}{4} \\ \frac34 \end{pmatrix},$

we get

$P\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix} \frac{\sqrt3}{4} \\ -\frac14 \end{pmatrix}.$

Also,

$T\begin{pmatrix}0\\1\end{pmatrix} = P\begin{pmatrix}0\\1\end{pmatrix} + \begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix} \frac14 \\ 1+\frac{\sqrt3}{4} \end{pmatrix},$

$P\begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix} \frac14 \\ \frac{\sqrt3}{4} \end{pmatrix}.$

Therefore the columns of $P$ are these two vectors:

$P= \begin{pmatrix} \frac{\sqrt3}{4} & \frac14 \\ -\frac14 & \frac{\sqrt3}{4} \end{pmatrix}$

📊 Marks: M1 A1

(b) Matrix $S$

An enlargement with scale factor $\frac12$ , centre $O$ , is represented by

$S= \begin{pmatrix} \frac12 & 0 \\ 0 & \frac12 \end{pmatrix}$

📊 Marks: A1

(c.i) Find $R$

Since $P=RS$ ,

$R=PS^{-1}$

and

$S^{-1}= \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}.$

Therefore

$R= \begin{pmatrix} \frac{\sqrt3}{4} & \frac14 \\ -\frac14 & \frac{\sqrt3}{4} \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} \frac{\sqrt3}{2} & \frac12 \\ -\frac12 & \frac{\sqrt3}{2} \end{pmatrix}$

📊 Marks: M1 A1

(c.ii) Angle and direction of the rotation

Comparing

$R= \begin{pmatrix} \frac{\sqrt3}{2} & \frac12 \\ -\frac12 & \frac{\sqrt3}{2} \end{pmatrix}$

with the standard rotation form, this is a rotation of

$30^\circ$

in the

$\text{clockwise direction about }O.$

📊 Marks: A1 A1

(d.i) Equation satisfied by $\begin{pmatrix}a\\b\end{pmatrix}$

Since $(a,b)$ is the centre, it is invariant under the transformation, so

$P\begin{pmatrix}a\\b\end{pmatrix} + \begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}a\\b\end{pmatrix}$

📊 Marks: M1

(d.ii) Find $a$ and $b$

Solve

$\begin{pmatrix} \frac{\sqrt3}{4} & \frac14 \\ -\frac14 & \frac{\sqrt3}{4} \end{pmatrix} \begin{pmatrix}a\\b\end{pmatrix} + \begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}a\\b\end{pmatrix}.$

This gives

$a=\frac{5+2\sqrt3}{13}, \qquad b=\frac{14+3\sqrt3}{13}.$

So the centre is

$\left(\frac{5+2\sqrt3}{13},\frac{14+3\sqrt3}{13}\right)$

📊 Marks: M1 A1 A1

⚠ Common Mistakes

Forgetting that the image of the origin gives $\mathbf{q}$ directly.
Using the transformed coordinates as the columns of $P$ without first subtracting $\mathbf{q}$ .
Mixing up clockwise and anticlockwise rotation matrices.
Using $R=P^{-1}S$ instead of $R=PS^{-1}$ .
Solving for the centre without using the invariant-point equation.

📘 IB Exam Tips

Always find $\mathbf{q}$ first when the origin is one of the given points.
Use the images of $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ to build the matrix $P$ .
When factoring a matrix, think geometrically before multiplying anything out.
For the centre of a transformation, write the invariant equation before solving.
Keep exact surd values until the final answer.

🧪 Challenge Problem

A transformation $U$ is given by

$U(\mathbf{x})=M\mathbf{x}+\mathbf{t},$

where the points

$O(0,0),\qquad A(1,0),\qquad B(0,1)$

are transformed to

$O'(1,2),\qquad A'(2,2),\qquad B'(1,3).$

(a) Find $\mathbf{t}$ .

(b) Find the matrix $M$ .

(c) Describe fully the transformation represented by $M$ .

(d) Find the invariant point of $U$ .

✅ Self-Practice

A transformation $V$ is given by

$V(\mathbf{x})=N\mathbf{x}+\mathbf{s},$

where

$O(0,0),\qquad A(1,0),\qquad B(0,1)$

are transformed to

$O'(0,2),\qquad A'(0,3),\qquad B'(-1,2).$

(a) Find $\mathbf{s}$ .

(b) Find the matrix $N$ .

(c) Describe fully the transformation represented by $N$ .

(d) Find the invariant point of $V$ .

Show Solutions

Challenge Problem

(a)

$\mathbf{t}= \begin{pmatrix} 1 \\ 2 \end{pmatrix}$

(b)

Since

$U\begin{pmatrix}1\\0\end{pmatrix}= \begin{pmatrix}2\\2\end{pmatrix}, \qquad U\begin{pmatrix}0\\1\end{pmatrix}= \begin{pmatrix}1\\3\end{pmatrix},$

subtract $\mathbf{t}$ :

$M\begin{pmatrix}1\\0\end{pmatrix}= \begin{pmatrix}1\\0\end{pmatrix}, \qquad M\begin{pmatrix}0\\1\end{pmatrix}= \begin{pmatrix}0\\1\end{pmatrix}.$

$M= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

(c)

$M$ is the identity transformation.

(d)

Solve

$\mathbf{x}+\begin{pmatrix}1\\2\end{pmatrix}=\mathbf{x},$

which has no solution.

So $U$ has no invariant point.

Self-Practice

(a)

$\mathbf{s}= \begin{pmatrix} 0 \\ 2 \end{pmatrix}$

(b)

Subtract $\mathbf{s}$ from the images of the basis vectors:

$N\begin{pmatrix}1\\0\end{pmatrix}= \begin{pmatrix}0\\1\end{pmatrix}, \qquad N\begin{pmatrix}0\\1\end{pmatrix}= \begin{pmatrix}-1\\0\end{pmatrix}.$

Hence

$N= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$

(c)

$N$ represents a rotation of $90^\circ$ anticlockwise about the origin.

(d)

Solve

$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix} + \begin{pmatrix}0\\2\end{pmatrix} = \begin{pmatrix}x\\y\end{pmatrix}.$

This gives

$-y=x,\qquad x+2=y.$

$x=-1,\qquad y=1.$

Therefore the invariant point is

$(-1,1).$

Vector Transformations and Composite Matrices

Lesson Overview

⭐ Key Concepts

📘 Clear IB-Style Explanation

📌 Worked Example — Affine Transformation of a Triangle

Solution

⚠ Common Mistakes

📘 IB Exam Tips

🧪 Challenge Problem

✅ Self-Practice

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Vector Transformations 2

Vector Transformations and Composite Matrices

Lesson Overview

⭐ Key Concepts

📘 Clear IB-Style Explanation

📌 Worked Example — Affine Transformation of a Triangle

Solution

⚠ Common Mistakes

📘 IB Exam Tips

🧪 Challenge Problem

✅ Self-Practice

Related

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