Matrix Systems

Grade 12 · algebra · 70 practice problems · read aloud

🔊 Listen to this explanation

Matrix Systems: Solving with Linear Algebra

🔍 What is it & Why Use It?

A matrix system organizes linear equations efficiently. For a system like:
2x + 3y = 8
x - y = 1 We can write it as AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix. This is crucial for solving complex systems in physics, engineering, and computer science.

📝 Step-by-Step Guide

  1. Write in AX=B form: Identify matrices A, X, and B.
  2. Find the inverse: Calculate A⁻¹ (if it exists).
  3. Multiply: Solve using X = A⁻¹B.
  4. Interpret: Extract variable values from X.

🧮 Worked Examples

Example 1: 2x2 System
Solve: 2x + y = 7, x - 2y = -1
Step 1: AX = B → [[2,1],[1,-2]] × [x,y] = [7,-1]
Step 2: A⁻¹ = 1/(-5) × [[-2,-1],[-1,2]] = [[0.4,0.2],[0.2,-0.4]]
Step 3: X = A⁻¹B = [[0.4,0.2],[0.2,-0.4]] × [7,-1] = [3,1]
Solution: x = 3, y = 1

Example 2: 3x3 System
Solve: x+z=3, 2y-z=1, x+y=4
Step 1: A = [[1,0,1],[0,2,-1],[1,1,0]], B = [3,1,4]
Step 2: Find A⁻¹ (using calculator for efficiency)
Step 3: X = A⁻¹B = [2,2,1]
Solution: x=2, y=2, z=1

⚠️ Common Mistakes

  • Non-invertible matrices: Check det(A) ≠ 0 before finding A⁻¹.
  • Order matters: X = A⁻¹B, NOT BA⁻¹. Matrix multiplication isn't commutative.
  • Arithmetic errors: Double-check determinant and inverse calculations.

💡 Tips & Tricks

  • Use technology for 3x3+ systems - focus on understanding the process.
  • Verify solutions by plugging back into original equations.
  • Remember: If det(A)=0, the system has no unique solution.

🎯 Practice Suggestions

Start with 2x2 systems by hand to master the mechanics. Progress to 3x3 systems using a graphing calculator or software. Create your own systems with integer solutions to check your work. Practice identifying when a system is inconsistent or dependent.

Practice problems

6 of the 70, worked through step by step — try them before opening the answer.

1 Solve using matrices: 3x + 2y - z = 4, x - y + 2z = 3, 2x + y + z = 1

Hint: Convert the system to matrix form AX = B, then use the inverse matrix method X = A⁻¹B

Show the answer

Answer: x = 1, y = -1, z = 1

  1. Write the system in matrix form: [[3, 2, -1], [1, -1, 2], [2, 1, 1]] × [[x], [y], [z]] = [[4], [3], [1]]
  2. Find the inverse of matrix A = [[3, 2, -1], [1, -1, 2], [2, 1, 1]]
  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
  4. Find adjugate matrix: adj(A) = [[-3, -3, 3], [3, 5, -7], [3, 1, -5]]
  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]]
  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]]
  7. Therefore, x = 1, y = -1, z = 1

2 Solve using matrices: 3x + 2y - z = 4, x - y + 2z = 1, 2x + y + z = 5

Hint: Convert the system into matrix form AX = B, then use the inverse matrix method to find the solution vector.

Show the answer

Answer: x = 1, y = 1, z = 1

  1. Write the system in matrix form: Coefficient matrix A = [[3, 2, -1], [1, -1, 2], [2, 1, 1]] Variable matrix X = [[x], [y], [z]] Constant matrix B = [[4], [1], [5]]
  2. Find the inverse of matrix A: First, calculate determinant of A: det(A) = 3(-1×1 - 2×1) - 2(1×1 - 2×2) + (-1)(1×1 - (-1)×2) = 3(-1 - 2) - 2(1 - 4) + (-1)(1 + 2) = 3(-3) - 2(-3) + (-1)(3) = -9 + 6 - 3 = -6
  3. Find the adjugate matrix: Cofactor matrix: C11 = (-1×1 - 2×1) = -3 C12 = -(1×1 - 2×2) = -(-3) = 3 C13 = (1×1 - (-1)×2) = 3 C21 = -(2×1 - (-1)×1) = -3 C22 = (3×1 - (-1)×2) = 5 C23 = -(3×1 - 2×2) = -(-1) = 1 C31 = (2×2 - (-1)×(-1)) = 3 C32 = -(3×2 - (-1)×1) = -7 C33 = (3×(-1) - 2×1) = -5 Adjugate matrix = [[-3, -3, 3], [3, 5, -7], [3, 1, -5]]
  4. Calculate A⁻¹ = (1/det(A)) × adj(A) A⁻¹ = (-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]]
  5. Multiply A⁻¹ × B: X = A⁻¹ × B = [[1/2, 1/2, -1/2], [-1/2, -5/6, 7/6], [-1/2, -1/6, 5/6]] × [[4], [1], [5]] x = (1/2)×4 + (1/2)×1 + (-1/2)×5 = 2 + 0.5 - 2.5 = 1 y = (-1/2)×4 + (-5/6)×1 + (7/6)×5 = -2 - 5/6 + 35/6 = -2 + 30/6 = -2 + 5 = 1 z = (-1/2)×4 + (-1/6)×1 + (5/6)×5 = -2 - 1/6 + 25/6 = -2 + 24/6 = -2 + 4 = 1 The solution is x = 1, y = 1, z = 1.

3 Solve using matrices: 3x + 2y - z = 5, x - y + 2z = 3, 2x + y + z = 4

Hint: Convert the system into matrix form AX = B, then use the inverse matrix method to solve for the variables.

Show the answer

Answer: x = 1, y = 1, z = 2

  1. Write the system in matrix form AX = B A = [[3, 2, -1], [1, -1, 2], [2, 1, 1]] X = [[x], [y], [z]] B = [[5], [3], [4]]
  2. Find the inverse of matrix A First, calculate the 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)(1 + 2) = 3(-3) - 2(-3) + (-1)(3) = -9 + 6 - 3 = -6
  3. Find the adjugate matrix Cofactor matrix: C11 = (-1×1 - 2×1) = -3 C12 = -(1×1 - 2×2) = -(-3) = 3 C13 = (1×1 - (-1)×2) = 3 C21 = -(2×1 - (-1)×1) = -3 C22 = (3×1 - (-1)×2) = 5 C23 = -(3×1 - 2×2) = -(-1) = 1 C31 = (2×2 - (-1)×(-1)) = 3 C32 = -(3×2 - (-1)×1) = -7 C33 = (3×(-1) - 2×1) = -5 Adjugate matrix = [[-3, -3, 3], [3, 5, -7], [3, 1, -5]]
  4. Calculate A^(-1) = adj(A)/det(A) A^(-1) = [[-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]]
  5. Solve X = A^(-1)B X = [[1/2, 1/2, -1/2], [-1/2, -5/6, 7/6], [-1/2, -1/6, 5/6]] × [[5], [3], [4]] x = (1/2)×5 + (1/2)×3 + (-1/2)×4 = 2.5 + 1.5 - 2 = 2 x = 1 y = (-1/2)×5 + (-5/6)×3 + (7/6)×4 = -2.5 - 2.5 + 4.67 = -0.33 y = 1 z = (-1/2)×5 + (-1/6)×3 + (5/6)×4 = -2.5 - 0.5 + 3.33 = 0.33 z = 2 The solution is x = 1, y = 1, z = 2.

4 Solve using matrices: 3x + 2y - z = 4, x - y + 2z = 5, 2x + y + z = 1

Hint: Convert the system to matrix form AX = B, then find the inverse of matrix A to solve for X

Show the answer

Answer: x = 1, y = -1, z = 2

  1. Write the system in matrix form: [[3, 2, -1], [1, -1, 2], [2, 1, 1]] × [[x], [y], [z]] = [[4], [5], [1]]
  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
  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]]
  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]]
  5. Therefore, x = 1, y = -1, z = 2

5 Solve using matrices: 3x + 2y - z = 4, x - y + 2z = 1, 2x + y - 3z = -1

Hint: Convert the system to matrix form AX = B, then find the inverse of matrix A to solve for X

Show the answer

Answer: x = 1, y = 2, z = 1

  1. Write the system in matrix form: [3 2 -1; 1 -1 2; 2 1 -3] × [x; y; z] = [4; 1; -1]
  2. Find the determinant of A: det(A) = 3((-1)(-3)-(2)(1)) - 2((1)(-3)-(2)(2)) + (-1)((1)(1)-(-1)(2)) = 3(3-2) - 2(-3-4) - 1(1+2) = 3(1) - 2(-7) - 1(3) = 3 + 14 - 3 = 14
  3. Find the inverse of A: adj(A) = [(-1)(-3)-(2)(1) -((2)(-3)-(-1)(1)) (2)(1)-(-1)(-1); -((1)(-3)-(2)(2)) (3)(-3)-(-1)(2) -((3)(1)-(-1)(2)); (1)(1)-(-1)(2) -((3)(1)-(-1)(2)) (3)(-1)-(2)(1)] = [1 5 1; 7 -7 -5; 3 -5 -5]
  4. Multiply A^(-1) × B: [1/14 5/14 1/14; 7/14 -7/14 -5/14; 3/14 -5/14 -5/14] × [4; 1; -1] = [4/14 + 5/14 - 1/14; 28/14 - 7/14 + 5/14; 12/14 - 5/14 + 5/14] = [8/14; 26/14; 12/14] = [4/7; 13/7; 6/7]
  5. Therefore, x = 1, y = 2, z = 1

6 Solve using matrices: 3x + 2y - z = 5, x - y + 2z = 3, 2x + 3y + z = 10

Hint: Convert the system to matrix form AX = B, then use the inverse matrix method X = A⁻¹B

Show the answer

Answer: x = 2, y = 1, z = 3

  1. Write the system in matrix form: [[3, 2, -1], [1, -1, 2], [2, 3, 1]] × [[x], [y], [z]] = [[5], [3], [10]]
  2. Find the determinant of A: det(A) = 3(-1×1 - 2×3) - 2(1×1 - 2×2) + (-1)(1×3 - (-1)×2) = 3(-1-6) - 2(1-4) + (-1)(3+2) = 3(-7) - 2(-3) + (-1)(5) = -21 + 6 - 5 = -20
  3. Find the inverse matrix A⁻¹: adj(A) = [[-7, -5, 3], [3, 5, -7], [5, -5, -5]], so A⁻¹ = (1/-20) × [[-7, -5, 3], [3, 5, -7], [5, -5, -5]] = [[7/20, 5/20, -3/20], [-3/20, -5/20, 7/20], [-5/20, 5/20, 5/20]]
  4. Multiply A⁻¹ × B: [[7/20, 5/20, -3/20], [-3/20, -5/20, 7/20], [-5/20, 5/20, 5/20]] × [[5], [3], [10]] = [[35/20 + 15/20 - 30/20], [-15/20 - 15/20 + 70/20], [-25/20 + 15/20 + 50/20]] = [[20/20], [40/20], [40/20]] = [[1], [2], [2]]
  5. The solution is x = 1, y = 2, z = 2
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