Matrix Transformations

The columns of the matrix are the images of (1,\ 0) and (0,\ 1).

1. A reflection in y=x swaps the coordinates of every point.
2. (1,\ 0)\to(0,\ 1), so the first column is \begin{pmatrix}0\\1\end{pmatrix}
3. (0,\ 1)\to(1,\ 0), so the second column is \begin{pmatrix}1\\0\end{pmatrix}
4. (\begin{pmatrix}0&-1\\-1&0\end{pmatrix} is a reflection in y=-x; \begin{pmatrix}1&0\\0&1\end{pmatrix} leaves every point where it is.)
5. Answer: \begin{pmatrix}0&1\\1&0\end{pmatrix}

The columns of the matrix are the images of (1, 0) and (0, 1).

1. First column: (1, 0) maps to (0, -1)
2. Second column: (0, 1) maps to (1, 0)
3. Both points have turned a quarter turn clockwise about O.
4. Rotation 90^\circ clockwise about the origin

Work out where the enlargement sends (1,\ 0) and (0,\ 1): each moves half as far from the origin.

1. (1,\ 0)\to(\frac{1}{2},\ 0): first column \begin{pmatrix}\frac{1}{2}\\0\end{pmatrix}
2. (0,\ 1)\to(0,\ \frac{1}{2}): second column \begin{pmatrix}0\\\frac{1}{2}\end{pmatrix}
3. (\begin{pmatrix}2&0\\0&2\end{pmatrix} is scale factor 2; \begin{pmatrix}\frac{1}{2}&0\\0&1\end{pmatrix} only halves the x-coordinates.)
4. Answer: \begin{pmatrix}\frac{1}{2}&0\\0&\frac{1}{2}\end{pmatrix}

The columns of the matrix are the images of (1,\ 0) and (0,\ 1): see which one moves.

1. First column: (1,\ 0)\to(-1,\ 0)
2. Second column: (0,\ 1)\to(0,\ 1), so points on the y-axis stay where they are.
3. In general (x,\ y)\to(-x,\ y): the x-coordinate changes sign.
4. Answer: reflection in the y-axis (the line x=0)

Find \mathbf{R} from the images of (1,\ 0) and (0,\ 1), then multiply it by the column vector of the point.

1. (1,\ 0)\to(0,\ 1) and (0,\ 1)\to(-1,\ 0), so \mathbf{R}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}
2. \begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}-3\\-5\end{pmatrix}
3. Top: 0(-3)+(-1)(-5)=5
4. Bottom: 1(-3)+0(-5)=-3
5. Answer: (5,\ -3)

Find \mathbf{R} by working out where (1, 0) and (0, 1) go, then write \mathbf{R}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}3\\-7\end{pmatrix}.

1. (1, 0)\to(0, -1) and (0, 1)\to(-1, 0), so \mathbf{R}=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}
2. Let P=(x, y): \;\begin{pmatrix}0&-1\\-1&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}-y\\-x\end{pmatrix}
3. -y=3 so y=-3
4. -x=-7 so x=7
5. P=(7, -3)

A is (1,\ 0) and C is (0,\ 1), so their images give the two columns of \mathbf{M}, in that order.

1. A(1,\ 0)\to A'(0,\ -3), so the first column is \begin{pmatrix}0\\-3\end{pmatrix}
2. C(0,\ 1)\to C'(3,\ 0), so the second column is \begin{pmatrix}3\\0\end{pmatrix}
3. Check with B: \begin{pmatrix}0&3\\-3&0\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}3\\-3\end{pmatrix}, which is B' ✓
4. (Writing the images as rows gives \begin{pmatrix}0&-3\\3&0\end{pmatrix}, which is wrong.)
5. Answer: \begin{pmatrix}0&3\\-3&0\end{pmatrix}

Write down the matrix for each reflection and multiply them, with the second reflection's matrix on the left.

1. Reflection in the x-axis: \begin{pmatrix}1&0\\0&-1\end{pmatrix}
2. Reflection in the y-axis: \begin{pmatrix}-1&0\\0&1\end{pmatrix}
3. Second on the left: \begin{pmatrix}-1&0\\0&1\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}
4. (1,\ 0)\to(-1,\ 0) and (0,\ 1)\to(0,\ -1): a half turn.
5. Answer: rotation 180^\circ about the origin

Work out \mathbf{R}^2=\mathbf{R}\mathbf{R} first, then multiply by \mathbf{R} again.

1. \mathbf{R}^2=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}0&-1\\1&0\end{pmatrix}=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}
2. \mathbf{R}^3=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\begin{pmatrix}0&-1\\1&0\end{pmatrix}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}
3. (1,\ 0)\to(0,\ -1) and (0,\ 1)\to(1,\ 0): a quarter turn clockwise.
4. (Also: \mathbf{R} is 90^\circ anticlockwise, so \mathbf{R}^3 is 270^\circ anticlockwise, the same as 90^\circ clockwise.)
5. Answer: rotation 90^\circ clockwise about the origin

(3,\ 0) is 3 times (1,\ 0), so its image is 3 times the image of (1,\ 0).

1. (3,\ 0)\to(0,\ -3), so (1,\ 0)\to(0,\ -1)
2. (0,\ 2)\to(2,\ 0), so (0,\ 1)\to(1,\ 0)
3. These images are the columns of \mathbf{T}.
4. Check: \begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}3\\0\end{pmatrix}=\begin{pmatrix}0\\-3\end{pmatrix} ✓
5. Answer: \begin{pmatrix}0&1\\-1&0\end{pmatrix}

Write down the two matrices for each pair and see whether \mathbf{PQ}=\mathbf{QP}.

1. Order matters when \mathbf{PQ}\ne\mathbf{QP}.
2. Rotation 90^\circ clockwise is \begin{pmatrix}0&1\\-1&0\end{pmatrix}; the enlargement is \begin{pmatrix}3&0\\0&3\end{pmatrix}=3\mathbf{I}
3. Both orders give \begin{pmatrix}0&3\\-3&0\end{pmatrix}, because 3\mathbf{I} just multiplies every entry by 3.
4. For example, \begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}0&-1\\1&0\end{pmatrix}=\begin{pmatrix}0&-1\\-1&0\end{pmatrix} but \begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}=\begin{pmatrix}0&1\\1&0\end{pmatrix}
5. Answer: rotation 90^\circ clockwise about the origin and enlargement, scale factor 3, centre the origin

Apply \mathbf{K} to B first, then apply \mathbf{L} to the result (or use the single matrix \mathbf{LK}).

1. \mathbf{K} first, so the combined matrix is \mathbf{LK} (second transformation on the left).
2. \mathbf{LK}=\begin{pmatrix}-1&0\\0&1\end{pmatrix}\begin{pmatrix}0&1\\-1&0\end{pmatrix}=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}
3. \begin{pmatrix}0&-1\\-1&0\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}-1\\-1\end{pmatrix}
4. (Step by step: \mathbf{K} sends (1,\ 1) to (1,\ -1), then \mathbf{L} sends that to (-1,\ -1).)
5. Answer: B'(-1,\ -1)

Write down \mathbf{P} and \mathbf{Q}, then multiply with the matrix of the second transformation on the left.

1. \mathbf{P}=\begin{pmatrix}0&-1\\-1&0\end{pmatrix} and \mathbf{Q}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}
2. \mathbf{P} first, so the combined matrix is \mathbf{QP}.
3. \mathbf{QP}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}0&-1\\-1&0\end{pmatrix}
4. Top row: 0(0)+1(-1)=-1 and 0(-1)+1(0)=0
5. Bottom row: -1(0)+0(-1)=0 and -1(-1)+0(0)=1
6. (\mathbf{PQ}=\begin{pmatrix}1&0\\0&-1\end{pmatrix} is the wrong order.)
7. Answer: \begin{pmatrix}-1&0\\0&1\end{pmatrix}

Let \mathbf{X}=\begin{pmatrix}a&b\\c&d\end{pmatrix}; the reflection happens first, so \mathbf{X} goes on the left of the product.

1. Reflection in the x-axis: \begin{pmatrix}1&0\\0&-1\end{pmatrix}; rotation: \begin{pmatrix}0&-1\\1&0\end{pmatrix}
2. \mathbf{X} is second, so \mathbf{X}\begin{pmatrix}1&0\\0&-1\end{pmatrix}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}
3. With \mathbf{X}=\begin{pmatrix}a&b\\c&d\end{pmatrix}: \;\begin{pmatrix}a&-b\\c&-d\end{pmatrix}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}
4. So a=0, b=1, c=1, d=0: \mathbf{X}=\begin{pmatrix}0&1\\1&0\end{pmatrix}
5. Answer: reflection in the line y=x

Multiply the two matrices to get the combined matrix in terms of k, then apply it to (a,\ 3).

1. Reflection in y=x: \begin{pmatrix}0&1\\1&0\end{pmatrix}; enlargement: \begin{pmatrix}k&0\\0&k\end{pmatrix}
2. Combined: \begin{pmatrix}k&0\\0&k\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}0&k\\k&0\end{pmatrix}
3. \begin{pmatrix}0&k\\k&0\end{pmatrix}\begin{pmatrix}a\\3\end{pmatrix}=\begin{pmatrix}3k\\ka\end{pmatrix}=\begin{pmatrix}12\\20\end{pmatrix}
4. 3k=12, so k=4
5. 4a=20
6. Answer: a=5

Write down the matrix for a reflection in y=-x and match the four entries: each one gives an equation. Answer part (a) before attempting part (b).

Part (a)

1. Reflection in y=-x: (1,\ 0)\to(0,\ -1) and (0,\ 1)\to(-1,\ 0), so the matrix is \begin{pmatrix}0&-1\\-1&0\end{pmatrix}
2. Top left: p^2-9=0, so p=3 or p=-3
3. Bottom right: q^2-4=0, so q=2 or q=-2
4. Top right: p+q=-1, which works only for p=-3,\ q=2
5. Bottom left: pq+5=-6+5=-1 ✓
6. Answer: p=-3

Part (b)

1. From part (a), p+q=-1 with p=-3
2. Answer: q=2

Take out a common factor of 3 from \mathbf{T} and look at the matrix that is left.

Part (a)

1. \mathbf{T}=3\begin{pmatrix}0&-1\\-1&0\end{pmatrix}
2. \begin{pmatrix}0&-1\\-1&0\end{pmatrix} is a reflection in y=-x, and 3\mathbf{I}=\begin{pmatrix}3&0\\0&3\end{pmatrix} is an enlargement, scale factor 3
3. Check: \begin{pmatrix}0&-1\\-1&0\end{pmatrix}\begin{pmatrix}3&0\\0&3\end{pmatrix}=\begin{pmatrix}0&-3\\-3&0\end{pmatrix} ✓
4. Answer: enlargement, scale factor 3, centre the origin, followed by a reflection in y=-x

Part (b)

1. \begin{pmatrix}0&-3\\-3&0\end{pmatrix}\begin{pmatrix}5\\4\end{pmatrix}
2. Top: 0(5)+(-3)(4)=-12
3. Bottom: -3(5)+0(4)=-15
4. Answer: B'(-12,\ -15)

Part (c)

1. A': \;\begin{pmatrix}0&-3\\-3&0\end{pmatrix}\begin{pmatrix}1\\2\end{pmatrix}=\begin{pmatrix}-6\\-3\end{pmatrix}
2. Gradient of A'B': \dfrac{-15-(-3)}{-12-(-6)}=\dfrac{-12}{-6}=2
3. y=2x+c through (-6,\ -3): -3=-12+c, so c=9
4. Check with B': 2(-12)+9=-15 ✓
5. Answer: y=2x+9

Use the rotation matrix to write P' in terms of t, then use Pythagoras' theorem for the distance PP'. Answer part (a) before attempting part (b).

Part (a)

1. Rotation 90^\circ clockwise: \begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}t\\2\end{pmatrix}=\begin{pmatrix}2\\-t\end{pmatrix}, so P'=(2,\ -t)
2. PP'^2=(t-2)^2+(2-(-t))^2
3. =t^2-4t+4+t^2+4t+4=2t^2+8
4. 2t^2+8=40, so t^2=16
5. t>0
6. Answer: t=4

Part (b)

1. P'=(2,\ -t) with t=4
2. Answer: P'(2,\ -4)

Part (c)

1. A rotation of 90^\circ about O keeps the distance from O the same and turns OP through a right angle.
2. So triangle OPP' has a right angle at O and OP=OP'
3. OP^2=4^2+2^2=20
4. Area =\frac{1}{2}\times OP\times OP'=\frac{1}{2}\times20
5. Answer: 10

Write down both matrices by tracking (1, 0) and (0, 1). The transformation that happens second goes on the LEFT when you multiply.

Part (a)

1. (1, 0)\to(0, -1) and (0, 1)\to(1, 0), so \mathbf{A}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}
2. \mathbf{B}=\begin{pmatrix}1&0\\0&-1\end{pmatrix}
3. \mathbf{A} first, so the combined matrix is \mathbf{BA} (second transformation on the left).
4. \mathbf{BA}=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}0&1\\-1&0\end{pmatrix}=\begin{pmatrix}0&1\\1&0\end{pmatrix}
5. \begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}5\\-2\end{pmatrix}=\begin{pmatrix}-2\\5\end{pmatrix}
6. Image is (-2, 5)

Part (b)

1. \mathbf{BA}=\begin{pmatrix}0&1\\1&0\end{pmatrix} swaps the coordinates: (1, 0)\to(0, 1) and (0, 1)\to(1, 0)
2. Reflection in the line y=x
3. (Reflection in y=-x comes from multiplying in the wrong order, \mathbf{AB}.)

Part (c)

1. The enlargement matrix is \begin{pmatrix}2&0\\0&2\end{pmatrix}
2. Overall matrix: \;\begin{pmatrix}2&0\\0&2\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}0&2\\2&0\end{pmatrix}
3. Let Q=(x, y): \;\begin{pmatrix}0&2\\2&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}2y\\2x\end{pmatrix}=\begin{pmatrix}8\\-6\end{pmatrix}
4. 2y=8 so y=4; 2x=-6 so x=-3
5. Q=(-3, 4)

A reflection keeps the area the same, but an enlargement with scale factor k multiplies the area by k^2. Answer part (a) before attempting part (b).

Part (a)

1. UV=3 and UW=4, with a right angle at U
2. Area of UVW=\frac{1}{2}\times3\times4=6
3. A reflection does not change area; an enlargement, scale factor k, multiplies area by k^2
4. 6k^2=150, so k^2=25
5. k>0
6. Answer: k=5

Part (b)

1. \mathbf{M}=\begin{pmatrix}5&0\\0&5\end{pmatrix}\begin{pmatrix}0&-1\\-1&0\end{pmatrix}=\begin{pmatrix}0&-5\\-5&0\end{pmatrix}
2. \begin{pmatrix}0&-5\\-5&0\end{pmatrix}\begin{pmatrix}1\\5\end{pmatrix}=\begin{pmatrix}-25\\-5\end{pmatrix}
3. Answer: W'(-25,\ -5)

Part (c)

1. \mathbf{M}^2=\begin{pmatrix}0&-5\\-5&0\end{pmatrix}\begin{pmatrix}0&-5\\-5&0\end{pmatrix}=\begin{pmatrix}25&0\\0&25\end{pmatrix}
2. (Reflecting twice in the same line puts every point back, so only the two enlargements are left: 5\times5=25)
3. Answer: enlargement, scale factor 25, centre the origin