Applications: Matrices as Transformations

A matrix transformation is fixed by what it does to the two unit vectors i and j: their images are the two columns of the matrix. Once the matrix is known, every point of the shape is moved by writing it as a column and multiplying it by the matrix.

1. A quarter turn counterclockwise sends i, which points along the x-axis, to 01, up the y-axis. It sends j, which points up the y-axis, to −10, along the negative x-axis.
2. (a) The columns of the matrix are the images of i and j, in that order, so R = 0−110.
3. Write the corner A as a column and multiply it by R: R11 = 0 × 1 − 1 × 11 × 1 + 0 × 1 = −11, so A' is (−1, 1).
4. In the same way, R41 = 0 − 14 + 0 = −14 and R13 = 0 − 31 + 0 = −31.
5. (b) The rotated logo has corners A'(−1, 1), B'(−1, 4) and C'(−3, 1). Check: AB runs 4 − 1 = 3 cm across, and A'B' runs 4 − 1 = 3 cm up, so the side keeps its length of 3 cm, as it must under a rotation.

answer(a) i goes to 01 and j to −10, so R = 0−110; (b) A'(−1, 1), B'(−1, 4) and C'(−3, 1), and A'B' = AB = 3 cm

techniqueMatrices as Transformations of the Plane

The determinant of a 2 × 2 matrix is the factor by which the transformation multiplies every area. So the area of an image can be found in two ways: directly from the image's own corners, or as the area of the original times the determinant. The two must agree.

1. The old bed has parallel sides PQ = 5 − 1 = 4 m and SR = 4 − 2 = 2 m, and they are 5 − 3 = 2 m apart. Its area is 12(4 + 2) × 2 = 6 square meters.
2. Multiply each corner by M. The stretch multiplies every x-coordinate by 3 and every y-coordinate by 2: M13 = 36, M53 = 156, M45 = 1210 and M25 = 610.
3. (a) The new bed has corners P'(3, 6), Q'(15, 6), R'(12, 10) and S'(6, 10). Its parallel sides are 15 − 3 = 12 m and 12 − 6 = 6 m, and they are 10 − 6 = 4 m apart, so its area is 12(12 + 6) × 4 = 36 square meters.
4. The determinant is det M = 3 × 2 − 0 × 0 = 6, so the stretch multiplies every area by 6.
5. (b) The old area times the determinant is 6 × 6 = 36 square meters, which agrees with the area found directly. The bed is 3 times as wide and 2 times as long, and 3 × 2 = 6.

answer(a) P'(3, 6), Q'(15, 6), R'(12, 10) and S'(6, 10), and the new bed has an area of 36 square meters; (b) det M = 6, and 6 × 6 = 36 square meters

techniqueThe Determinant as an Area Scale Factor · Matrices as Transformations of the Plane

The size of the determinant is the factor by which areas are multiplied. Its sign says whether the shape keeps its orientation: a negative determinant turns the shape over, so corners that ran counterclockwise round the shape run clockwise round the image.

1. Fxy = 0 × x + 1 × y1 × x + 0 × y = yx, so F swaps the two coordinates of every point.
2. (a) A' is (1, 4), B' is (1, 7) and C' is (3, 4). Each corner and its image are the same distance from the mirror line y = x, on opposite sides of it.
3. The determinant is det F = 0 × 0 − 1 × 1 = −1. Its size is 1, so the area does not change: the stencil has area 12 × 3 × 2 = 3 square decimeters, and the image, with sides A'B' = 3 and A'C' = 2 at a right angle, also has area 3 square decimeters.
4. Its sign is negative, so the stencil has been turned over. Going from A to B to C round the stencil is counterclockwise, and going from A' to B' to C' round the image is clockwise.
5. (b) det F = −1: the area stays 3 square decimeters, and the minus sign shows that the image is the stencil turned over, a mirror image in which the order of the corners is reversed.

answer(a) A'(1, 4), B'(1, 7) and C'(3, 4); (b) det F = −1: the area of 3 square decimeters is unchanged, and the minus sign shows the stencil is turned over, with its corners now running clockwise

techniqueThe Determinant as an Area Scale Factor · Matrices as Transformations of the Plane

When a point p is transformed by R and then by X, it becomes X(Rp) = (XR)p. So the single matrix is the product with the first transformation on the RIGHT, read from right to left. Matrix multiplication is not commutative, so the other order can give a different result.

1. The turn is done first, so its matrix stands next to the point: the single matrix is XR, read from right to left.
2. XR = 100−10−110 = 1 × 0 + 0 × 11 × (−1) + 0 × 00 × 0 + (−1) × 10 × (−1) + (−1) × 0 = 0−1−10.
3. Check with one corner in two stages. The turn sends A(2, 1) to A'(−1, 2), and the reflection sends A' to A''(−1, −2). In one step, XR21 = −1−2, the same point.
4. (a) XR sends (x, y) to (−y, −x), so the tile ends up at A''(−1, −2), B''(−1, −4) and C''(−4, −2). The single matrix is the reflection in the line y = −x.
5. (b) The other order is RX = 0−110100−1 = 0110, which sends (x, y) to (y, x). The tile ends up at (1, 2), (1, 4) and (4, 2), its reflection in y = x, in a different place, because XR ≠ RX.

answer(a) XR = 0−1−10, and the tile ends up at (−1, −2), (−1, −4) and (−4, −2); (b) RX = 0110, which puts the tile at (1, 2), (1, 4) and (4, 2) instead

techniqueComposing Transformations by Multiplying Matrices · Matrices as Transformations of the Plane

An invariant point is one the transformation leaves where it is: Sp = p. An invariant line is a line whose points all land on the same line, though they may move along it. A line of invariant points is invariant in the stronger sense that no point on it moves at all.

1. Sxy = x + yy: each point slides sideways by its height y and keeps its height.
2. (a) A and B are on the table, where y = 0, so they stay at (1, 0) and (5, 0). C(5, 3) goes to C'(5 + 3, 3) = (8, 3), and D(1, 3) goes to D'(1 + 3, 3) = (4, 3).
3. The pushed stack is a parallelogram with base AB = 4 cm and height 3 cm, so its area is 4 × 3 = 12 square centimeters, the same as the rectangle's. This agrees with det S = 1 × 1 − 1 × 0 = 1: the shear keeps every area.
4. An invariant point satisfies Sxy = xy, so x + y = x and y = y. The first equation gives y = 0, so every point on the table line y = 0 is an invariant point.
5. (b) A point (x, c) on the line y = c goes to (x + c, c), which is on the same line, so every line y = c is an invariant line: each card slides along its own level. Only on y = 0 does every point stay where it is.

answer(a) A(1, 0), B(5, 0), C'(8, 3) and D'(4, 3), and the area is still 12 square centimeters; (b) the invariant points are the points of the table line y = 0, and every line y = c is an invariant line, though only y = 0 is a line of invariant points

techniqueInvariant Points and Lines · The Determinant as an Area Scale Factor

The invariant points solve Mp = p. A line y = mx through the origin is invariant when the image of a point on it lies on it again, which gives an equation for m. A line can be invariant without being a line of invariant points, when its points slide along it.

1. Multiply each corner by M: M11 = 33, M−1−1 = −3−3, M−11 = −2 + 1−1 + 2 = −11 and M1−1 = 1−1. So P' is (3, 3) and R' is (−3, −3), while Q and S do not move.
2. An invariant point satisfies Mxy = xy, so 2x + y = x and x + 2y = y. Both equations simplify to x + y = 0.
3. (a) The line of invariant points is y = −x, the pinned diagonal through Q and S.
4. A point (x, mx) on the line y = mx goes to (2x + mx, x + 2mx). The image is on the same line when x + 2mx = m(2x + mx). Divide by x: 1 + 2m = 2m + m², so m² = 1.
5. So m = 1 or m = −1. On y = x each point moves three times as far from the center, as P(1, 1) goes to (3, 3), so the line is invariant but its points move along it. On y = −x no point moves at all.
6. (b) The invariant lines through the origin are y = x and y = −x. det M = 2 × 2 − 1 × 1 = 3, so the fabric's area of 2 × 2 = 4 square decimeters becomes 3 × 4 = 12 square decimeters. Check: the diagonals of the stretched fabric are 6√2 and 2√2, and 12 × 6√2 × 2√2 = 12.

answer(a) P'(3, 3), Q'(−1, 1), R'(−3, −3) and S'(1, −1), and the line of invariant points is y = −x; (b) the invariant lines through the origin are y = x and y = −x, and the stretched fabric has an area of 12 square decimeters

techniqueInvariant Points and Lines · The Determinant as an Area Scale Factor

A 3 × 3 matrix transforms space, and its three columns are the images of the unit vectors i, j and k. Its determinant is the factor by which it multiplies every volume, just as a 2 × 2 determinant multiplies every area.

1. The columns of M are the images of i, j and k: i goes to 300, j to 020 and k to 002. So M stretches the length by 3, and the depth and the height by 2.
2. (a) M212 = 3 × 22 × 12 × 2 = 624, so P' is (6, 2, 4).
3. Expand the determinant along the first row. Only its first entry is not 0: det M = 3 × (2 × 2 − 0 × 0) − 0 + 0 = 3 × 4 = 12.
4. The old shed has a volume of 2 × 1 × 2 = 4 cubic meters, and M multiplies every volume by 12, so the new shed has a volume of 12 × 4 = 48 cubic meters.
5. (b) det M = 12, and the new shed has a volume of 48 cubic meters. Check: the new shed runs from O to P'(6, 2, 4), so it is 6 m long, 2 m deep and 4 m high, and 6 × 2 × 4 = 48 cubic meters.

answer(a) i, j and k go to 300, 020 and 002, and P' is (6, 2, 4); (b) det M = 12, so the new shed has a volume of 48 cubic meters

techniqueThe Determinant as a Volume Scale Factor · Matrices as Transformations of Space

A turn about the z-axis moves i and j round in the horizontal plane and leaves k, which points along the axle, where it is. The three images are the columns of the matrix. The determinant gives the volume factor, and the invariant points solve Rp = p.

1. Seen from above, the quarter turn sends i to j and sends j to −i, and k points along the axle and does not move. These images are the columns: R = 0−10100001.
2. (a) R411 = 0 × 4 − 1 × 1 + 01 × 4 + 0 + 00 + 0 + 1 × 1 = −141, so P' is (−1, 4, 1). The corner has swung round the axle and kept its height of 1 m.
3. Expand the determinant along the third row, whose first two entries are 0: det R = 1 × (0 × 0 − (−1) × 1) = 1. So the turn multiplies every volume by 1, and the crate keeps its volume of 2 × 1 × 1 = 2 cubic meters.
4. An invariant point satisfies Rxyz = xyz, so −y = x, x = y and z = z. Together the first two give x = −x, so x = 0 and then y = 0, while z can be any number.
5. (b) det R = 1, so the crate's volume of 2 cubic meters is unchanged. The invariant points are the points (0, 0, z) of the axle: they are the only points the turn does not move.

answer(a) R = 0−10100001 and P'(−1, 4, 1); (b) det R = 1, so the crate keeps its volume of 2 cubic meters, and the invariant points are the points (0, 0, z) on the axle

techniqueMatrices as Transformations of Space · The Determinant as a Volume Scale Factor · Invariant Points and Lines

A determinant of 0 multiplies every area by 0, so the whole plane is squashed onto a line (or a point). Then different points land on the same image, and no transformation can send one image back to two places, so the matrix has no inverse.

1. Mxy = x + yx + y, so every point goes to a point whose two coordinates are equal.
2. (a) A' is (4, 4), B' is (6, 6), C' is (7, 7) and D' is (5, 5). The determinant is det M = 1 × 1 − 1 × 1 = 0.
3. Every image lies on the line y = x, so the photo is flattened onto the segment of that line from (4, 4) to (7, 7).
4. The photo's area of 2 × 1 = 2 square centimeters is multiplied by det M = 0, so the image has an area of 2 × 0 = 0, as a piece of a line must.
5. (b) Different points land on the same point: B(5, 1) and the point (4, 2) on the top edge both go to (6, 6), because 5 + 1 = 4 + 2. A matrix that undid M would have to send (6, 6) back to two places at once, so none exists; det M = 0 says the same, because a matrix with determinant 0 has no inverse.

answer(a) A'(4, 4), B'(6, 6), C'(7, 7) and D'(5, 5), and det M = 0; (b) the whole photo is flattened onto the line y = x with an area of 0, and since points such as (5, 1) and (4, 2) land on the same point, no matrix can undo it

techniqueThe Determinant as an Area Scale Factor · Matrices as Transformations of the Plane

A rotation through θ counterclockwise about the origin sends i to cos θsin θ and j to −sin θcos θ, so these are the columns of its matrix. A rotation keeps every length, and its determinant is cos² θ + sin² θ = 1.

1. The columns are the images of i and j: R = cos θ−sin θsin θcos θ = 35−454535.
2. Multiply the first two corners by R: R55 = 3 − 44 + 3 = −17 and R105 = 6 − 48 + 3 = 211.
3. (a) In the same way C(10, 10) goes to (6 − 8, 8 + 6) = (−2, 14) and D(5, 10) goes to (3 − 8, 4 + 6) = (−5, 10). The turned bed has corners A'(−1, 7), B'(2, 11), C'(−2, 14) and D'(−5, 10).
4. From A' to B' is 2 − (−1) = 3 m across and 11 − 7 = 4 m up, so by Pythagoras A'B' = √3² + 4² = √25 = 5 m, the same as AB = 10 − 5 = 5 m.
5. (b) A'B' = 5 m, and det R = 35 × 35 − (−45) × 45 = 925 + 1625 = 1. So the bed keeps its side of 5 m and its area of 5 × 5 = 25 square meters: the turn moves the bed without changing its size or its shape.

answer(a) R = 35−454535, and the corners are A'(−1, 7), B'(2, 11), C'(−2, 14) and D'(−5, 10); (b) A'B' = 5 m, the same as AB, and det R = 1, so the area stays 25 square meters

techniqueMatrices as Transformations of the Plane · The Determinant as an Area Scale Factor