Matrix Equations Systems Worksheet Answer Key – Precalculus

Problem 1 Writing a Matrix Equation

The coefficients form A, the unknowns form X and the constants form B:

X is the column with entries x and y, and B is the column with entries 9 and −3.

Answer: A has rows (4, −1) and (2, 5), X = (x, y) as a column, B = (9, −3) as a column

Problem 2 Reading a Matrix Equation

The first row gives 2x − 3y = 8 and the second row gives 5x + y = 3.

Test (1, −2): 2(1) − 3(−2) = 2 + 6 = 8 and 5(1) + (−2) = 3. Both hold.

Answer: 2x − 3y = 8 and 5x + y = 3; yes, (1, −2) is a solution

Problem 3 Is There One Solution?

A has rows (6, 9) and (2, 3). det A = 6(3) − 9(2) = 18 − 18 = 0.

A zero determinant means no unique solution. The left sides are proportional (the first is 3 times the second) but 5 ≠ 3 · 7, so the lines are parallel and there is no solution.

Answer: det = 0, so there is no unique solution (the lines are parallel)

Problem 4 Matrix Times a Point

First entry: 3(5) + (−1)(−2) = 15 + 2 = 17. Second entry: 2(5) + 4(−2) = 10 − 8 = 2.

Answer: the column (17, 2)

Problem 5 Reflecting a Point

First entry: −1(6) + 0(−4) = −6. Second entry: 0(6) + 1(−4) = −4.

Answer: (−6, −4)

Problem 6 Stretching the Unit Square

det = 3(2) − 0(0) = 6. The image is a 3 by 2 rectangle with area 6 square units, which equals |det|.

Answer: det = 6; area 6 square units

Problem 7 Solving with an Inverse

A has rows (4, 3) and (5, 2); det A = 4(2) − 3(5) = −7. So A⁻¹ = −17 times rows (2, −3) and (−5, 4).

X = A⁻¹B = −17(2·6 − 3·11, −5·6 + 4·11) = −17(−21, 14) = (3, −2).

Check: 4(3) + 3(−2) = 6 and 5(3) + 2(−2) = 11.

Answer: x = 3 and y = −2

Problem 8 A Negative Solution

det A = 3(−1) − 4(2) = −11. A⁻¹ = −111 times rows (−1, −4) and (−2, 3).

X = −111((−1)(−7) + (−4)(−12), (−2)(−7) + 3(−12)) = −111(55, −22) = (−5, 2).

Check: 3(−5) + 4(2) = −7 and 2(−5) − 2 = −12.

Answer: x = −5 and y = 2

Problem 9 A Fractional Solution

det A = 2(−1) − 3(4) = −14. A⁻¹ = −114 times rows (−1, −3) and (−4, 2).

X = −114(−5 − 18, −20 + 12) = −114(−23, −8) = (2314, 47).

Check: 2(2314) + 3(47) = 237 + 127 = 5 and 4(2314) − 47 = 467 − 47 = 6.

Answer: x = 2314 and y = 47

Problem 10 Two Moves in a Row

Rotation: (0)(−3) + (−1)(5) = −5 and (1)(−3) + 0(5) = −3, giving (−5, −3). Reflection in the x-axis changes the sign of y: (−5, 3).

The combined matrix is the reflection matrix times the rotation matrix, in that order:

Applying it to (−3, 5): 0(−3) − 1(5) = −5 and −1(−3) + 0(5) = 3, which matches.

Answer: (−5, 3); combined matrix with rows (0, −1) and (−1, 0)

Problem 11 Area After a Transformation

det = 4(3) − (−2)(1) = 12 + 2 = 14. Areas are multiplied by |det| = 14.

New area = 6 · 14 = 84 square units.

Answer: 84 square units

Problem 12 Building a Scaling Matrix

The images of (1, 0) and (0, 1) are (2, 0) and (0, 3), so the matrix has rows (2, 0) and (0, 3).

Image of (2, −1): (2·2 + 0, 0 + 3(−1)) = (4, −3). Determinant = 6, so the unit square becomes a rectangle of area 6.

Answer: matrix with rows (2, 0), (0, 3); image (4, −3); area 6

Problem 13 A Matrix That Flattens

det = 2(6) − 4(3) = 0. The image of (1, 0) is (2, 3) and the image of (0, 1) is (4, 6).

Both images lie on the line y = 1.5x. Every point is sent to that line, so the plane collapses onto a line and areas become 0. The matrix has no inverse.

Answer: det = 0; the plane collapses onto the line y = 1.5x

Problem 14 A Shear

(0, 0) stays (0, 0). (3, 0) goes to (3 + 0, 0) = (3, 0). (0, 2) goes to (0 + 2·2, 2) = (4, 2).

The original area is 12(3)(2) = 3. The image has base 3 and height 2, so its area is also 3. This agrees with det = 1(1) − 2(0) = 1.

Answer: images (0, 0), (3, 0), (4, 2); both areas equal 3

Problem 15 Food Truck Prices

Let w and b be the prices. A has rows (18, 12) and (25, 20), and B is the column (246, 365).

det A = 18(20) − 12(25) = 360 − 300 = 60, so A⁻¹ = 160 times rows (20, −12) and (−25, 18).

w = 20·246 − 12·36560 = 54060 = 9 and b = −25·246 + 18·36560 = 42060 = 7.

Check: 18(9) + 12(7) = 246 and 25(9) + 20(7) = 365.

Answer: a wrap costs $9 and a bowl costs $7

Problem 16 Mixing Solutions

Total volume: x + y = 12. Acid: 0.2x + 0.5y = 0.38(12) = 4.56. A has rows (1, 1) and (0.2, 0.5), B = (12, 4.56).

det A = 1(0.5) − 1(0.2) = 0.3. A⁻¹ = 10.3 times rows (0.5, −1) and (−0.2, 1).

x = 0.5(12) − 4.560.3 = 1.440.3 = 4.8 and y = −0.2(12) + 4.560.3 = 2.160.3 = 7.2.

Check: 4.8 + 7.2 = 12 and 0.2(4.8) + 0.5(7.2) = 0.96 + 3.6 = 4.56.

Answer: 4.8 liters of the 20% solution and 7.2 liters of the 50% solution

Problem 17 Theater Ticket Prices

Dividing each equation by 5 gives 8a + 6c + 2s = 162, 5a + 9c + 4s = 168 and 6a + 4c + 6s = 158. A has rows (8, 6, 2), (5, 9, 4), (6, 4, 6) and B is the column (162, 168, 158).

Technology gives det A = 200 and A⁻¹ = 1200 times the matrix:

Multiplying by B gives the column (12, 8, 9).

Check: 40(12) + 30(8) + 10(9) = 810; 25(12) + 45(8) + 20(9) = 840; 30(12) + 20(8) + 30(9) = 790.

Answer: adult $12, child $8, senior $9

Problem 18 Robot Arm Zone

(2, 1) goes to (0(2) − 2.5(1), 2.5(2) + 0) = (−2.5, 5). (6, 1) goes to (−2.5, 15). (2, 4) goes to (−10, 5).

Original area: 12(4)(3) = 6. det = 0(0) − (−2.5)(2.5) = 6.25, so the new area is 6(6.25) = 37.5.

Direct check: the new triangle has a vertical side from (−2.5, 5) to (−2.5, 15) of length 10 and a horizontal distance of 7.5 to the vertex (−10, 5), so its area is 12(10)(7.5) = 37.5.

Answer: vertices (−2.5, 5), (−2.5, 15), (−10, 5); area 37.5 square units

Problem 19 Transformation from Images

(a) The images become the columns: M has rows (3, −1) and (1, 2).

(b) (3(2) − 1(5), 1(2) + 2(5)) = (1, 12).

(c) det = 3(2) − (−1)(1) = 7. Areas are multiplied by 7, and orientation is preserved because det > 0.

(d) M⁻¹ = 17 times rows (2, 1) and (−1, 3). Applying it to (8, 5): 17(16 + 5, −8 + 15) = 17(21, 7) = (3, 1). Check: M(3, 1) = (9 − 1, 3 + 2) = (8, 5).

Answer: (a) rows (3, −1), (1, 2); (b) (1, 12); (c) det = 7; (d) (3, 1)

Problem 20 When Is There No Unique Solution?

(a) det = 2(6) − k(3) = 12 − 3k = 0 gives k = 4.

(b) With k = 4: 2x + 4y = 8 simplifies to x + 2y = 4, and 3x + 6y = 12 simplifies to the same equation. The lines coincide, so there are infinitely many solutions.

(c) For k = 5, det = 12 − 15 = −3 and A⁻¹ = −13 times rows (6, −5) and (−3, 2). X = −13(6·8 − 5·12, −3·8 + 2·12) = −13(−12, 0) = (4, 0).

Check: 2(4) + 5(0) = 8 and 3(4) + 6(0) = 12.

Answer: (a) k = 4; (b) infinitely many solutions; (c) (4, 0)