Matrices as transformations Problem set B One question at a time

Find the image of each point under the transformation with the given matrix.

1. (a) \(\begin{pmatrix}3 & -1 \\ 2 & 5\end{pmatrix}\), point \((2, 3)\)

   • \(\begin{pmatrix}3 & -1 \\ 2 & 5\end{pmatrix}\begin{pmatrix}2 \\ 3\end{pmatrix} = \begin{pmatrix}3 \\ 19\end{pmatrix}\), so the image is \((3, 19)\).

2. (b) \(\begin{pmatrix}2 & 1 \\ -3 & 4\end{pmatrix}\), point \((-3, 2)\)

   • \(\begin{pmatrix}2 & 1 \\ -3 & 4\end{pmatrix}\begin{pmatrix}-3 \\ 2\end{pmatrix} = \begin{pmatrix}-4 \\ 17\end{pmatrix}\), so the image is \((-4, 17)\).

3. (c) \(\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\), point \((-6, 2)\)

   • \(\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\begin{pmatrix}-6 \\ 2\end{pmatrix} = \begin{pmatrix}2 \\ 6\end{pmatrix}\), so the image is \((2, 6)\).

Write down the matrix for each transformation.

1. (a) Horizontal stretch, scale factor 4

   • Horizontal stretch: \((1,0)\) goes to \((4,0)\) and \((0,1)\) stays at \((0,1)\). These are the columns.
   • \(M = \begin{pmatrix}4 & 0 \\ 0 & 1\end{pmatrix}\)

2. (b) Vertical stretch, scale factor \(\frac{1}{2}\)

   • Vertical stretch: \((1,0)\) stays at \((1,0)\) and \((0,1)\) goes to \((0,\frac{1}{2})\). These are the columns.
   • \(M = \begin{pmatrix}1 & 0 \\ 0 & \frac{1}{2}\end{pmatrix}\)

3. (c) Enlargement, scale factor −3

   • Enlargement with scale factor \(-3\): \((1,0)\) goes to \((-3,0)\) and \((0,1)\) goes to \((0,-3)\).
   • \(M = \begin{pmatrix}-3 & 0 \\ 0 & -3\end{pmatrix}\)

4. (d) Rotation 90° anticlockwise about the origin

   • Rotation by \(\theta\): \(\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}\) with \(\theta = 90^\circ\).
   • \(M = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\)

5. (e) Rotation 45° clockwise about the origin

   • Rotation by \(\theta\): \(\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}\) with \(\theta = -45^\circ\) (clockwise).
   • \(M = \begin{pmatrix}\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{pmatrix}\)

6. (f) Reflection in the line \(y = \frac{1}{\sqrt{3}}\,x\)

   • The line makes angle \(\theta\) with the \(x\)-axis where \(\tan\theta = \frac{1}{\sqrt{3}}\), so \(\theta = 30^\circ\). Reflection: \(\begin{pmatrix}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta\end{pmatrix}\) with \(2\theta = 60^\circ\).
   • \(M = \begin{pmatrix}\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2}\end{pmatrix}\)

Find the point whose image is given.

1. (a) \(\begin{pmatrix}3 & -2 \\ 1 & 4\end{pmatrix}\), image \((4, 6)\)

   • \(\begin{pmatrix}3 & -2 \\ 1 & 4\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}4 \\ 6\end{pmatrix}\)
   • \(\det = (3)(4) - (-2)(1) = 14\), so \(\begin{pmatrix}x \\ y\end{pmatrix} = \frac{1}{14}\begin{pmatrix}4 & 2 \\ -1 & 3\end{pmatrix}\begin{pmatrix}4 \\ 6\end{pmatrix} = \begin{pmatrix}2 \\ 1\end{pmatrix}\).

2. (b) \(\begin{pmatrix}2 & 1 \\ -1 & 3\end{pmatrix}\), image \((6, 11)\)

   • \(\begin{pmatrix}2 & 1 \\ -1 & 3\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}6 \\ 11\end{pmatrix}\)
   • \(\det = (2)(3) - (1)(-1) = 7\), so \(\begin{pmatrix}x \\ y\end{pmatrix} = \frac{1}{7}\begin{pmatrix}3 & -1 \\ 1 & 2\end{pmatrix}\begin{pmatrix}6 \\ 11\end{pmatrix} = \begin{pmatrix}1 \\ 4\end{pmatrix}\).

The unit square has vertices \(O(0,0),\ A(1,0),\ C(1,1),\ B(0,1)\). Find the vertices of its image under each matrix, and sketch it.

1. (a) \(\begin{pmatrix}2 & 1 \\ 1 & 3\end{pmatrix}\)

   • O stays at \((0,0)\). \(A(1,0)\) goes to the first column, \((2,1)\), and \(B(0,1)\) goes to the second column, \((1,3)\).
   • \(C(1,1)\) goes to \(\begin{pmatrix}2 & 1 \\ 1 & 3\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix} = \begin{pmatrix}3 \\ 4\end{pmatrix}\), so the image is the parallelogram \(O,\ (2,1),\ (3,4),\ (1,3)\).

2. (b) \(\begin{pmatrix}0 & 3 \\ -3 & 0\end{pmatrix}\)

   • O stays at \((0,0)\). \(A(1,0)\) goes to the first column, \((0,-3)\), and \(B(0,1)\) goes to the second column, \((3,0)\).
   • \(C(1,1)\) goes to \(\begin{pmatrix}0 & 3 \\ -3 & 0\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix} = \begin{pmatrix}3 \\ -3\end{pmatrix}\), so the image is the parallelogram \(O,\ (0,-3),\ (3,-3),\ (3,0)\).

3. (c) \(\begin{pmatrix}1 & -1 \\ 2 & 1\end{pmatrix}\)

   • O stays at \((0,0)\). \(A(1,0)\) goes to the first column, \((1,2)\), and \(B(0,1)\) goes to the second column, \((-1,1)\).
   • \(C(1,1)\) goes to \(\begin{pmatrix}1 & -1 \\ 2 & 1\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix} = \begin{pmatrix}0 \\ 3\end{pmatrix}\), so the image is the parallelogram \(O,\ (1,2),\ (0,3),\ (-1,1)\).

Each transformation sends \((1,0)\) and \((0,1)\) to the points given. Write down its matrix.

1. (a) \((5, -1)\) and \((2, 3)\)

   • The images of \((1,0)\) and \((0,1)\) are the columns of the matrix.
   • \(M = \begin{pmatrix}5 & 2 \\ -1 & 3\end{pmatrix}\)

2. (b) \((0, -4)\) and \((4, 0)\)

   • The images of \((1,0)\) and \((0,1)\) are the columns of the matrix.
   • \(M = \begin{pmatrix}0 & 4 \\ -4 & 0\end{pmatrix}\)

3. (c) \((2, 0)\) and \((-3, 1)\)

   • The images of \((1,0)\) and \((0,1)\) are the columns of the matrix.
   • \(M = \begin{pmatrix}2 & -3 \\ 0 & 1\end{pmatrix}\)

Describe the transformation with each matrix.

1. (a) \(\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\)

   • Columns \((0,1)\) and \((1,0)\): \(x\) and \(y\) swap, which is a reflection in \(y = x\).

2. (b) \(\begin{pmatrix}-3 & 0 \\ 0 & -3\end{pmatrix}\)

   • Every point goes to \(-3\) times its position: an enlargement with scale factor \(-3\) (a rotation of 180° and an enlargement with scale factor 3).

3. (c) \(\begin{pmatrix}5 & 0 \\ 0 & 5\end{pmatrix}\)

   • Every point goes to 5 times its position: an enlargement with scale factor 5.

4. (d) \(\begin{pmatrix}0.8 & -0.6 \\ 0.6 & 0.8\end{pmatrix}\)

   • The form is \(\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}\), so \(\cos\theta = 0.8\) and \(\sin\theta = 0.6\).
   • \(\theta = \tan^{-1}\left(\frac{0.6}{0.8}\right) = 36.9^\circ\) anticlockwise.

(a) Find the matrix for a reflection in the line \(y = 2x\). (b) Find the point whose image under this reflection is \((10, 5)\).

• (a) \(\tan\theta = 2\) gives \(\cos 2\theta = \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \frac{1 - 4}{1 + 4} = -0.6\) and \(\sin 2\theta = \frac{2\tan\theta}{1 + \tan^2\theta} = \frac{4}{5} = 0.8\).
• \(M = \begin{pmatrix}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta\end{pmatrix} = \begin{pmatrix}-0.6 & 0.8 \\ 0.8 & 0.6\end{pmatrix}\).
• (b) \(M^2 = I\), so \(M\) is its own inverse and the required point is the image of \((10, 5)\).
• \(\begin{pmatrix}-0.6 & 0.8 \\ 0.8 & 0.6\end{pmatrix}\begin{pmatrix}10 \\ 5\end{pmatrix} = \begin{pmatrix}-2 \\ 11\end{pmatrix}\), so the point is \((-2, 11)\).