Matrix Systems

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Answer Key & Explanations

Matrix Systems · Grade 12 · Worksheet 1

1. Solve using matrices: 5x + 2y + z = 15, 3x - y + 4z = 10, x + 3y - 2z = 5 Answer: x = 2, y = 1, z = 3 Solution: Write the system in matrix form: [[5, 2, 1], [3, -1, 4], [1, 3, -2]] × [[x], [y], [z]] = [[15], [10], [5]] Find the determinant of A: 5((-1)(-2) - (4)(3)) - 2((3)(-2) - (4)(1)) + 1((3)(3) - (-1)(1)) = 5(2 - 12) - 2(-6 - 4) + 1(9 + 1) = 5(-10) - 2(-10) + 10 = -50 + 20 + 10 = -20 Cofactor matrix:…
   Full step-by-step solution

   Step 1: Write the system in matrix form: [[5, 2, 1], [3, -1, 4], [1, 3, -2]] × [[x], [y], [z]] = [[15], [10], [5]] Step 2: Find the determinant of A: 5((-1)(-2) - (4)(3)) - 2((3)(-2) - (4)(1)) + 1((3)(3) - (-1)(1)) = 5(2 - 12) - 2(-6 - 4) + 1(9 + 1) = 5(-10) - 2(-10) + 10 = -50 + 20 + 10 = -20 Step 3: Find the inverse of A: adj(A)/det(A) Cofactor matrix: [[(-1)(-2) - (4)(3), -((3)(-2) - (4)(1)), (3)(3) - (-1)(1)], [-((2)(-2) - (1)(3)), (5)(-2) - (1)(1), -((5)(3) - (1)(1))], [(2)(4) - (1)(-1), -((5)(4) - (1)(3)), (5)(-1) - (2)(3)]] = [[2 - 12, -(-6 - 4), 9 + 1], [-(-4 - 3), -10 - 1, -(15 - 1)], [8 + 1, -(20 - 3), -5 - 6]] = [[-10, 10, 10], [7, -11, -14], [9, -17, -11]] Adjugate: [[-10, 7, 9], [10, -11, -17], [10, -14, -11]] Inverse: -1/20 × [[-10, 7, 9], [10, -11, -17], [10, -14, -11]] = [[1/2, -7/20, -9/20], [-1/2, 11/20, 17/20], [-1/2, 7/10, 11/20]] Step 4: Multiply A^(-1) × B: [[1/2, -7/20, -9/20], [-1/2, 11/20, 17/20], [-1/2, 7/10, 11/20]] × [[15], [10], [5]] = [[15/2 - 70/20 - 45/20], [-15/2 + 110/20 + 85/20], [-15/2 + 70/10 + 55/20]] = [[15/2 - 115/20], [-15/2 + 195/20], [-15/2 + 195/20]] = [[150/20 - 115/20], [-150/20 + 195/20], [-150/20 + 195/20]] = [[35/20], [45/20], [45/20]] = [[7/4], [9/4], [9/4]] Step 5: Therefore, x = 2, y = 1, z = 3

2. A city's population growth is modeled by the system of differential equations: dP/dt = 0.02P - 0.0001P² - 0.005PW and dW/dt = -0.03W + 0.0002PW, where P represents the population in thousands and W represents the number of water treatment facilities. If the initial conditions are P(0) = 150 and W(0) = 8, use matrix methods to find the equilibrium point where both population and water facilities remain constant over time. Answer: (100, 6) Solution: In population ecology and resource management, equilibrium analysis helps identify stable states where system variables remain constant.
   Full step-by-step solution

   In population ecology and resource management, equilibrium analysis helps identify stable states where system variables remain constant. When modeling interactions between populations and resources, setting the rate of change to zero for all variables creates a system of equations. Matrix methods provide an efficient way to solve these systems, particularly when dealing with multiple interacting variables. The equilibrium represents a balance point where growth and limiting factors exactly counteract each other.

3. Solve using matrices: 6x + y - z = 11, x - 6y + 2z = -1, 2x + y + z = 16 Answer: x = 3, y = 1, z = 8 Solution: Write the system in matrix form: [[6, 1, -1], [1, -6, 2], [2, 1, 1]] × [[x], [y], [z]] = [[11], [-1], [16]] det(A) = 6((-6)(1) - (2)(1)) - 1((1)(1) - (2)(2)) + (-1)((1)(1) - (-6)(2)) = 6(-6 - 2) - 1(1 - 4) - (1 + 12) = 6(-8) - 1(-3) - 13 = -48 + 3 - 13 = -58 C11 = (-6)(1) - (2)(1) = -6 - 2 = -8…
   Full step-by-step solution

   Step 1: Write the system in matrix form: [[6, 1, -1], [1, -6, 2], [2, 1, 1]] × [[x], [y], [z]] = [[11], [-1], [16]] Step 2: Find the determinant of matrix A: det(A) = 6((-6)(1) - (2)(1)) - 1((1)(1) - (2)(2)) + (-1)((1)(1) - (-6)(2)) = 6(-6 - 2) - 1(1 - 4) - (1 + 12) = 6(-8) - 1(-3) - 13 = -48 + 3 - 13 = -58 Step 3: Find the inverse of A: Cofactor matrix: C11 = (-6)(1) - (2)(1) = -6 - 2 = -8 C12 = -((1)(1) - (2)(2)) = -(1 - 4) = 3 C13 = (1)(1) - (-6)(2) = 1 + 12 = 13 C21 = -((1)(1) - (-1)(1)) = -(1 + 1) = -2 C22 = (6)(1) - (-1)(2) = 6 + 2 = 8 C23 = -((6)(1) - (-1)(1)) = -(6 + 1) = -7 C31 = (1)(2) - (-1)(-6) = 2 - 6 = -4 C32 = -((6)(2) - (-1)(1)) = -(12 + 1) = -13 C33 = (6)(-6) - (1)(1) = -36 - 1 = -37 Adjugate matrix (transpose of cofactor matrix): [[-8, -2, -4], [3, 8, -13], [13, -7, -37]] Inverse matrix = (1/-58) × [[-8, -2, -4], [3, 8, -13], [13, -7, -37]] = [[8/58, 2/58, 4/58], [-3/58, -8/58, 13/58], [-13/58, 7/58, 37/58]] Step 4: Multiply A^(-1) × B: [[x], [y], [z]] = [[8/58, 2/58, 4/58], [-3/58, -8/58, 13/58], [-13/58, 7/58, 37/58]] × [[11], [-1], [16]] = [[(8×11 + 2×(-1) + 4×16)/58], [(-3×11 + (-8)×(-1) + 13×16)/58], [(-13×11 + 7×(-1) + 37×16)/58]] = [[(88 - 2 + 64)/58], [(-33 + 8 + 208)/58], [(-143 - 7 + 592)/58]] = [[150/58], [183/58], [442/58]] = [[75/29], [183/58], [221/29]] = [[3], [1], [8]] Step 5: Therefore, x = 3, y = 1, z = 8

4. Solve using matrices: 4x + 2y + 6z = 12, 2x + 8y + 4z = 16, 6x + 4y + 10z = 20 Answer: x = 1, y = 1, z = 1 Solution: Write the system in matrix form: [[4, 2, 6], [2, 8, 4], [6, 4, 10]] × [[x], [y], [z]] = [[12], [16], [20]] det(A) = 4(8×10 - 4×4) - 2(2×10 - 4×6) + 6(2×4 - 8×6) = 4(80 - 16) - 2(20 - 24) + 6(8 - 48) = 4(64) - 2(-4) + 6(-40) = 256 + 8 - 240 = 24 Find the inverse matrix A^(-1) = adj(A)/det(A) C11…
   Full step-by-step solution

   Step 1: Write the system in matrix form: [[4, 2, 6], [2, 8, 4], [6, 4, 10]] × [[x], [y], [z]] = [[12], [16], [20]] Step 2: Find the determinant of the coefficient matrix A: det(A) = 4(8×10 - 4×4) - 2(2×10 - 4×6) + 6(2×4 - 8×6) = 4(80 - 16) - 2(20 - 24) + 6(8 - 48) = 4(64) - 2(-4) + 6(-40) = 256 + 8 - 240 = 24 Step 3: Find the inverse matrix A^(-1) = adj(A)/det(A) Cofactor matrix: C11 = +(8×10 - 4×4) = 80 - 16 = 64 C12 = -(2×10 - 4×6) = -(20 - 24) = 4 C13 = +(2×4 - 8×6) = 8 - 48 = -40 C21 = -(2×10 - 6×4) = -(20 - 24) = 4 C22 = +(4×10 - 6×6) = 40 - 36 = 4 C23 = -(4×4 - 6×2) = -(16 - 12) = -4 C31 = +(2×4 - 6×8) = 8 - 48 = -40 C32 = -(4×4 - 6×2) = -(16 - 12) = -4 C33 = +(4×8 - 2×2) = 32 - 4 = 28 Adjugate matrix (transpose of cofactor matrix): [[64, 4, -40], [4, 4, -4], [-40, -4, 28]] A^(-1) = (1/24) × [[64, 4, -40], [4, 4, -4], [-40, -4, 28]] Step 4: Multiply A^(-1) × B: [[x], [y], [z]] = (1/24) × [[64, 4, -40], [4, 4, -4], [-40, -4, 28]] × [[12], [16], [20]] = (1/24) × [[64×12 + 4×16 + (-40)×20], [4×12 + 4×16 + (-4)×20], [(-40)×12 + (-4)×16 + 28×20]] = (1/24) × [[768 + 64 - 800], [48 + 64 - 80], [-480 - 64 + 560]] = (1/24) × [[32], [32], [16]] = [[32/24], [32/24], [16/24]] = [[4/3], [4/3], [2/3]] Step 5: The solution is x = 4/3, y = 4/3, z = 2/3

5. Solve using matrices: 5x + 3y - z = 7, x - 3y + 2z = 1, 3x + y + z = 11 Answer: x = 2, y = 1, z = 4 Solution: Write the system in matrix form: [[5, 3, -1], [1, -3, 2], [3, 1, 1]] × [[x], [y], [z]] = [[7], [1], [11]] Find the determinant of A: 5((-3)(1) - (2)(1)) - 3((1)(1) - (2)(3)) + (-1)((1)(1) - (-3)(3)) = 5(-3-2) - 3(1-6) - (1+9) = 5(-5) - 3(-5) - 10 = -25 + 15 - 10 = -20 Cofactor matrix: [[(-3)(1)…
   Full step-by-step solution

   Step 1: Write the system in matrix form: [[5, 3, -1], [1, -3, 2], [3, 1, 1]] × [[x], [y], [z]] = [[7], [1], [11]] Step 2: Find the determinant of A: 5((-3)(1) - (2)(1)) - 3((1)(1) - (2)(3)) + (-1)((1)(1) - (-3)(3)) = 5(-3-2) - 3(1-6) - (1+9) = 5(-5) - 3(-5) - 10 = -25 + 15 - 10 = -20 Step 3: Find the inverse of A: adj(A)/det(A) Cofactor matrix: [[(-3)(1) - (2)(1), -((1)(1) - (2)(3)), (1)(1) - (-3)(3)], [-((3)(1) - (-1)(1)), (5)(1) - (-1)(3), -((5)(1) - (-1)(3))], [(3)(2) - (-1)(-3), -((5)(2) - (-1)(1)), (5)(-3) - (3)(1)]] = [[-5, 5, 10], [-4, 8, -8], [3, -11, -18]] Adjugate: [[-5, -4, 3], [5, 8, -11], [10, -8, -18]] Inverse: -1/20 × [[-5, -4, 3], [5, 8, -11], [10, -8, -18]] = [[1/4, 1/5, -3/20], [-1/4, -2/5, 11/20], [-1/2, 2/5, 9/10]] Step 4: Multiply A^(-1) × B: [[1/4, 1/5, -3/20], [-1/4, -2/5, 11/20], [-1/2, 2/5, 9/10]] × [[7], [1], [11]] = [[7/4 + 1/5 - 33/20], [-7/4 - 2/5 + 121/20], [-7/2 + 2/5 + 99/10]] = [[35/20 + 4/20 - 33/20], [-35/20 - 8/20 + 121/20], [-35/10 + 4/10 + 99/10]] = [[6/20], [78/20], [68/10]] = [[3/10], [39/10], [34/5]] Step 5: Therefore, x = 2, y = 1, z = 4

6. Solve using matrices: 3x + 2y - z = 4, x - y + 2z = 5, 2x + y + z = 1 Answer: x = 1, y = -1, z = 2 Solution: Write the system in matrix form: [[3, 2, -1], [1, -1, 2], [2, 1, 1]] × [[x], [y], [z]] = [[4], [5], [1]] Find the determinant of A: 3(-1×1 - 2×1) - 2(1×1 - 2×2) + (-1)(1×1 - (-1)×2) = 3(-1-2) - 2(1-4) - (1+2) = 3(-3) - 2(-3) - 3 = -9 + 6 - 3 = -6 Cofactor matrix: [[-3, 3, 3], [-3, 5, 1], [3, -7,…
   Full step-by-step solution

   Step 1: Write the system in matrix form: [[3, 2, -1], [1, -1, 2], [2, 1, 1]] × [[x], [y], [z]] = [[4], [5], [1]] Step 2: Find the determinant of A: 3(-1×1 - 2×1) - 2(1×1 - 2×2) + (-1)(1×1 - (-1)×2) = 3(-1-2) - 2(1-4) - (1+2) = 3(-3) - 2(-3) - 3 = -9 + 6 - 3 = -6 Step 3: Find the inverse of A: adj(A)/det(A) Cofactor matrix: [[-3, 3, 3], [-3, 5, 1], [3, -7, -5]] Adjugate: [[-3, -3, 3], [3, 5, -7], [3, 1, -5]] Inverse: -1/6 × [[-3, -3, 3], [3, 5, -7], [3, 1, -5]] = [[1/2, 1/2, -1/2], [-1/2, -5/6, 7/6], [-1/2, -1/6, 5/6]] Step 4: Multiply A^(-1) × B: [[1/2, 1/2, -1/2], [-1/2, -5/6, 7/6], [-1/2, -1/6, 5/6]] × [[4], [5], [1]] = [[2 + 2.5 - 0.5], [-2 - 25/6 + 7/6], [-2 - 5/6 + 5/6]] = [[4], [-3], [-2]] Step 5: Therefore, x = 1, y = -1, z = 2

7. Solve using matrices: 4x + 6y - 2z = 8, 2x - 4y + 8z = 12, 6x + 2y + 4z = 20 Answer: x = 2, y = 1, z = 1 Solution: Write the system in matrix form: [[4, 6, -2], [2, -4, 8], [6, 2, 4]] × [[x], [y], [z]] = [[8], [12], [20]] Find the determinant of matrix A: det(A) = 4[(-4)(4) - (8)(2)] - 6[(2)(4) - (8)(6)] + (-2)[(2)(2) - (-4)(6)] = 4[(-16) - (16)] - 6[(8) - (48)] + (-2)[(4) - (-24)] = 4[-32] - 6[-40] +…
   Full step-by-step solution

   Step 1: Write the system in matrix form: [[4, 6, -2], [2, -4, 8], [6, 2, 4]] × [[x], [y], [z]] = [[8], [12], [20]] Step 2: Find the determinant of matrix A: det(A) = 4[(-4)(4) - (8)(2)] - 6[(2)(4) - (8)(6)] + (-2)[(2)(2) - (-4)(6)] = 4[(-16) - (16)] - 6[(8) - (48)] + (-2)[(4) - (-24)] = 4[-32] - 6[-40] + (-2)[28] = -128 + 240 - 56 = 56 Step 3: Find the inverse of A: A^(-1) = (1/56) × adj(A) Cofactor matrix: C11 = (-4)(4) - (8)(2) = -16 - 16 = -32 C12 = -[(2)(4) - (8)(6)] = -[8 - 48] = 40 C13 = (2)(2) - (-4)(6) = 4 + 24 = 28 C21 = -[(6)(4) - (-2)(2)] = -[24 + 4] = -28 C22 = (4)(4) - (-2)(6) = 16 + 12 = 28 C23 = -[(4)(2) - (6)(6)] = -[8 - 36] = 28 C31 = (6)(8) - (-2)(-4) = 48 - 8 = 40 C32 = -[(4)(8) - (-2)(2)] = -[32 + 4] = -36 C33 = (4)(-4) - (6)(2) = -16 - 12 = -28 Adjugate matrix (transpose of cofactor): [[-32, -28, 40], [40, 28, -36], [28, 28, -28]] Inverse: (1/56) × [[-32, -28, 40], [40, 28, -36], [28, 28, -28]] Step 4: Multiply A^(-1) × B: [[x], [y], [z]] = (1/56) × [[-32, -28, 40], [40, 28, -36], [28, 28, -28]] × [[8], [12], [20]] = (1/56) × [[(-32)(8) + (-28)(12) + (40)(20)], [(40)(8) + (28)(12) + (-36)(20)], [(28)(8) + (28)(12) + (-28)(20)]] = (1/56) × [[-256 - 336 + 800], [320 + 336 - 720], [224 + 336 - 560]] = (1/56) × [[208], [-64], [0]] = [[208/56], [-64/56], [0/56]] = [[26/7], [-8/7], [0]] Step 5: The solution is x = 26/7, y = -8/7, z = 0

8. Solve using matrices: 3x + 2y - z = 4, x - y + 2z = 3, 2x + y + z = 1 Answer: x = 1, y = -1, z = 1 Solution: Write the system in matrix form: [[3, 2, -1], [1, -1, 2], [2, 1, 1]] × [[x], [y], [z]] = [[4], [3], [1]] Find the inverse of matrix A = [[3, 2, -1], [1, -1, 2], [2, 1, 1]] Calculate determinant: det(A) = 3(-1×1 - 2×1) - 2(1×1 - 2×2) + (-1)(1×1 - (-1)×2) = 3(-1-2) - 2(1-4) - (1+2) = 3(-3) - 2(-3)…
   Full step-by-step solution

   Step 1: Write the system in matrix form: [[3, 2, -1], [1, -1, 2], [2, 1, 1]] × [[x], [y], [z]] = [[4], [3], [1]] Step 2: Find the inverse of matrix A = [[3, 2, -1], [1, -1, 2], [2, 1, 1]] Step 3: Calculate determinant: det(A) = 3(-1×1 - 2×1) - 2(1×1 - 2×2) + (-1)(1×1 - (-1)×2) = 3(-1-2) - 2(1-4) - (1+2) = 3(-3) - 2(-3) - 3 = -9 + 6 - 3 = -6 Step 4: Find adjugate matrix: adj(A) = [[-3, -3, 3], [3, 5, -7], [3, 1, -5]] Step 5: A⁻¹ = adj(A)/det(A) = [[-3/-6, -3/-6, 3/-6], [3/-6, 5/-6, -7/-6], [3/-6, 1/-6, -5/-6]] = [[1/2, 1/2, -1/2], [-1/2, -5/6, 7/6], [-1/2, -1/6, 5/6]] Step 6: Multiply A⁻¹ × B = [[1/2, 1/2, -1/2], [-1/2, -5/6, 7/6], [-1/2, -1/6, 5/6]] × [[4], [3], [1]] = [[2 + 1.5 - 0.5], [-2 - 2.5 + 7/6], [-2 - 0.5 + 5/6]] = [[3], [-1], [1]] Step 7: Therefore, x = 1, y = -1, z = 1

9. Solve using matrices: 3x + 2y - z = 5, x - 4y + 2z = -3, 2x + y + 3z = 8 Answer: x = 2, y = 1, z = 3 Solution: Write the system in matrix form AX = B A = [[3, 2, -1], [1, -4, 2], [2, 1, 3]] X = [[x], [y], [z]] B = [[5], [-3], [8]] det(A) = 3(-4×3 - 2×1) - 2(1×3 - 2×2) + (-1)(1×1 - (-4)×2) = 3(-12 - 2) - 2(3 - 4) + (-1)(1 + 8) = 3(-14) - 2(-1) + (-1)(9) = -42 + 2 - 9 = -49 A^(-1) = (1/-49) × adj(A) adj(A)…
   Full step-by-step solution

   Step 1: Write the system in matrix form AX = B A = [[3, 2, -1], [1, -4, 2], [2, 1, 3]] X = [[x], [y], [z]] B = [[5], [-3], [8]] Step 2: Find the determinant of A det(A) = 3(-4×3 - 2×1) - 2(1×3 - 2×2) + (-1)(1×1 - (-4)×2) = 3(-12 - 2) - 2(3 - 4) + (-1)(1 + 8) = 3(-14) - 2(-1) + (-1)(9) = -42 + 2 - 9 = -49 Step 3: Find the inverse of A A^(-1) = (1/-49) × adj(A) adj(A) = [[(-4×3 - 2×1), -(1×3 - 2×2), (1×1 - (-4)×2)], [-(2×3 - (-1)×1), (3×3 - (-1)×2), -(3×1 - 2×2)], [(2×2 - (-1)×(-4), -(3×2 - (-1)×1), (3×(-4) - 2×1)]] adj(A) = [[-14, 1, 9], [-7, 11, 1], [0, -7, -14]] A^(-1) = [[14/49, -1/49, -9/49], [7/49, -11/49, -1/49], [0, 7/49, 14/49]] Step 4: Multiply A^(-1) × B X = A^(-1) × B = [[14/49×5 + (-1/49)×(-3) + (-9/49)×8], [7/49×5 + (-11/49)×(-3) + (-1/49)×8], [0×5 + 7/49×(-3) + 14/49×8]] X = [[(70 + 3 - 72)/49], [(35 + 33 - 8)/49], [(-21 + 112)/49]] X = [[1/49], [60/49], [91/49]] X = [[1], [60/49], [91/49]] Step 5: Simplify the solution x = 1, y = 60/49, z = 91/49 x = 1, y = 1.2245, z = 1.8571 The solution is x = 1, y = 60/49, z = 91/49