A matrix moves points
- Write the point as a column vector and put the matrix in front: the image of \((x, y)\) is \(M\begin{pmatrix}x \\ y\end{pmatrix}\).
- To go back from an image to the point it came from, use \(M^{-1}\).
AHL 5.8
Matrices as transformations
AHL 5.8
Matrices as transformations
By the end of this section
- Find the image of a point, and the point with a given image
- Write down a matrix from where the unit square goes
- Use the matrices for rotations, reflections, stretches and enlargements
Key points
Read the matrix from the unit square
- The first column is where \((1, 0)\) goes. The second column is where \((0, 1)\) goes.
- So you can write down a matrix by asking where those two points land.
The standard matrices
- These are in the formula booklet: reflection in \(y = (\tan\theta)x\), horizontal and vertical stretches, enlargement, and rotations about the origin.
- Reflection in \(y = mx\): \(\tan\theta = m\), then use \(\cos 2\theta\) and \(\sin 2\theta\).
- Check whether a rotation is clockwise or anticlockwise before choosing the matrix.
On the calculator
Images of several points at once
m · (matrix of points)
- Put the points in as columns of one matrix. The product gives all the images, one in each column.
Screenshot and video to come
Explore: what a matrix does to the unit square
Drag the orange point to where (1, 0) goes: that is column 1 of M. Drag the teal point to where (0, 1) goes: column 2. Or type a matrix, or pick a transformation, and see where the points go.
M =
- (1, 0) goes to
- (0, 1) goes to
- Area of the image
- The square is
The points snap to halves. Select one (Tab) and use the arrow keys to move it.
- Drag the orange point to (0, 1) and the teal point to (−1, 0). Which transformation is it? Write down its matrix before you look at the boxes.
- Make the determinant 0. What happens to the square and to the grid? Why can this transformation not be undone?
- Which transformations turn the square over? What do their determinants have in common?
- Find two different matrices that give an image of area 6. Is there only one way?
Worked examples
Problems like the ones in class. Pick one, try it, then open the working.The image of a point
A transformation has matrix \(M = \begin{pmatrix}3 & -1 \\ 2 & 4\end{pmatrix}\). Find the image of the point \((2, 5)\).
Working
Have a go first. Then open the working one line at a time.
-
Multiply
\(\begin{pmatrix}3 & -1 \\ 2 & 4\end{pmatrix}\begin{pmatrix}2 \\ 5\end{pmatrix} = \begin{pmatrix}1 \\ 24\end{pmatrix}\)
The point goes after the matrix, as a column.
-
Answer
\((1, 24)\)
Read the column as a point.
Which point has this image?
For the same matrix \(M\), find the point whose image is \((7, 0)\).
Working
Have a go first. Then open the working one line at a time.
-
Use the inverse
\(\begin{pmatrix}x \\ y\end{pmatrix} = M^{-1}\begin{pmatrix}7 \\ 0\end{pmatrix}\)
The inverse undoes the transformation.
-
The inverse
\(M^{-1} = \frac{1}{14}\begin{pmatrix}4 & 1 \\ -2 & 3\end{pmatrix}\)
\(\det M = 14\). Or use the calculator.
-
Multiply
\(\frac{1}{14}\begin{pmatrix}4 & 1 \\ -2 & 3\end{pmatrix}\begin{pmatrix}7 \\ 0\end{pmatrix} = \begin{pmatrix}2 \\ -1\end{pmatrix}\)
The point is \((2, -1)\).
Find the matrix from the unit square
A transformation sends \((1, 0)\) to \((2, 1)\) and \((0, 1)\) to \((-1, 3)\). Write down its matrix.
Working
Have a go first. Then open the working one line at a time.
-
First column
\(\begin{pmatrix}2 \\ 1\end{pmatrix}\)
Where \((1, 0)\) goes.
-
Second column
\(\begin{pmatrix}-1 \\ 3\end{pmatrix}\)
Where \((0, 1)\) goes.
-
The matrix
\(\begin{pmatrix}2 & -1 \\ 1 & 3\end{pmatrix}\)
The two images, side by side as columns.
A rotation
Write down the matrix for a rotation of 90° anticlockwise about the origin, and find the image of \((4, 1)\).
Working
Have a go first. Then open the working one line at a time.
-
The matrix
\(\begin{pmatrix}\cos 90^\circ & -\sin 90^\circ \\ \sin 90^\circ & \cos 90^\circ\end{pmatrix} = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\)
From the formula booklet. Or: \((1, 0)\) goes to \((0, 1)\) and \((0, 1)\) goes to \((-1, 0)\).
-
The image
\(\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}4 \\ 1\end{pmatrix} = \begin{pmatrix}-1 \\ 4\end{pmatrix}\)
\((-1, 4)\).
A clockwise rotation
Write down the matrix for a rotation of 60° clockwise about the origin, and find the image of \((4, 2)\) to 3 significant figures.
Working
Have a go first. Then open the working one line at a time.
-
The matrix
\(\begin{pmatrix}\cos 60^\circ & \sin 60^\circ \\ -\sin 60^\circ & \cos 60^\circ\end{pmatrix} \approx \begin{pmatrix}0.5 & 0.866 \\ -0.866 & 0.5\end{pmatrix}\)
The clockwise matrix in the formula booklet. The minus sign is bottom left.
-
The image
\(\begin{pmatrix}\cos 60^\circ & \sin 60^\circ \\ -\sin 60^\circ & \cos 60^\circ\end{pmatrix}\begin{pmatrix}4 \\ 2\end{pmatrix} \approx \begin{pmatrix}3.73 \\ -2.46\end{pmatrix}\)
\((3.73, -2.46)\). Type the matrix with cos and sin in it, so nothing is rounded early.
-
Does it look right?
\((4, 2)\) is above the \(x\)-axis. The image is below it.
A clockwise turn of 60° takes it down past the \(x\)-axis.
A reflection in a sloping line
Find the matrix for a reflection in the line \(y = \frac{1}{2}x\), and the image of \((5, 0)\).
Working
Have a go first. Then open the working one line at a time.
-
The angle
\(\tan\theta = \frac{1}{2}\)
The gradient of the line is \(\tan\theta\).
-
cos 2θ and sin 2θ
\(\cos 2\theta = 0.6\), \(\sin 2\theta = 0.8\)
From the calculator: \(\theta = 0.4636\ldots\) radians.
-
The matrix
\(\begin{pmatrix}0.6 & 0.8 \\ 0.8 & -0.6\end{pmatrix}\)
\(\begin{pmatrix}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta\end{pmatrix}\).
-
The image
\(\begin{pmatrix}0.6 & 0.8 \\ 0.8 & -0.6\end{pmatrix}\begin{pmatrix}5 \\ 0\end{pmatrix} = \begin{pmatrix}3 \\ 4\end{pmatrix}\)
\((3, 4)\). It is still 5 from the origin.
A stretch
Write down the matrix for a horizontal stretch with scale factor 4, and find the image of \((-2, 3)\).
Working
Have a go first. Then open the working one line at a time.
-
The matrix
\(\begin{pmatrix}4 & 0 \\ 0 & 1\end{pmatrix}\)
Only the \(x\)-coordinates change.
-
The image
\((-8, 3)\)
\(4 \times (-2) = -8\). The \(y\)-coordinate stays 3.
An enlargement of a triangle
A triangle has vertices \(A(1, 1)\), \(B(4, 1)\) and \(C(2, 3)\). It is enlarged with scale factor 3, centre the origin. Write down the matrix and find the images of the three vertices.
Working
Have a go first. Then open the working one line at a time.
-
The matrix
\(\begin{pmatrix}3 & 0 \\ 0 & 3\end{pmatrix}\)
Both coordinates are multiplied by 3: \((1, 0)\) goes to \((3, 0)\) and \((0, 1)\) goes to \((0, 3)\).
-
All three points at once
\(\begin{pmatrix}3 & 0 \\ 0 & 3\end{pmatrix}\begin{pmatrix}1 & 4 & 2 \\ 1 & 1 & 3\end{pmatrix} = \begin{pmatrix}3 & 12 & 6 \\ 3 & 3 & 9\end{pmatrix}\)
Each column is a vertex. Each column of the answer is its image.
-
Answer
\(A'(3, 3)\), \(B'(12, 3)\), \(C'(6, 9)\)
Every image is 3 times as far from the origin.
Describe a transformation fully
Describe fully the transformation with matrix \(\begin{pmatrix}0.28 & -0.96 \\ 0.96 & 0.28\end{pmatrix}\).
Working
Have a go first. Then open the working one line at a time.
-
Determinant
\(0.28(0.28) - (-0.96)(0.96) = 1\)
Area stays the same and the shape is not turned over. That fits a rotation, not a reflection.
-
Compare with a rotation
\(\cos\theta = 0.28\) and \(\sin\theta = 0.96\)
It has the pattern \(\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}\), the anticlockwise rotation in the formula booklet.
-
The angle
\(\theta = 73.7^\circ\)
From \(\cos^{-1}(0.28)\). Both \(\cos\theta\) and \(\sin\theta\) are positive, so the angle is between 0° and 90°.
-
Describe fully
A rotation of 73.7° anticlockwise about the origin
Three things: the type, the angle with its direction, and the centre.
The matrix from two points and their images
A transformation with matrix \(M\) sends \((2, 1)\) to \((5, 5)\) and \((1, 3)\) to \((0, 10)\). Find \(M\).
Working
Have a go first. Then open the working one line at a time.
-
Points as columns
\(M\begin{pmatrix}2 & 1 \\ 1 & 3\end{pmatrix} = \begin{pmatrix}5 & 0 \\ 5 & 10\end{pmatrix}\)
The points are the columns on the left. Their images are the columns on the right, in the same order.
-
Which side?
\(M = \begin{pmatrix}5 & 0 \\ 5 & 10\end{pmatrix}\begin{pmatrix}2 & 1 \\ 1 & 3\end{pmatrix}^{-1}\)
The matrix of points is on the right of \(M\), so its inverse goes on the right.
-
The inverse
\(\begin{pmatrix}2 & 1 \\ 1 & 3\end{pmatrix}^{-1} = \frac{1}{5}\begin{pmatrix}3 & -1 \\ -1 & 2\end{pmatrix}\)
The determinant is \(2(3) - 1(1) = 5\).
-
Multiply
\(\begin{pmatrix}5 & 0 \\ 5 & 10\end{pmatrix}\cdot\frac{1}{5}\begin{pmatrix}3 & -1 \\ -1 & 2\end{pmatrix} = \begin{pmatrix}3 & -1 \\ 1 & 3\end{pmatrix}\)
Check: \(M\) sends \((2, 1)\) to \((5, 5)\).