Using matrices, solve x + y + z = 2, 2x − y + 3z = 8 and x + 2y − z = −1.

1. 1 · Write what is given and what is to be found. Given: three linear equations. To find: x, y and z by the matrix method.
2. 2 · Write the system as AX = B. A = [1, 1, 1; 2, −1, 3; 1, 2, −1], X = [x; y; z], B = [2; 8; −1].
3. 3 · Find |A|. |A| = 1(1 − 6) − 1(−2 − 3) + 1(4 + 1) = −5 + 5 + 5 = 5 ≠ 0, so A⁻¹ exists and the solution is unique.
4. 4 · Find the cofactors.
   • A₁₁ = −5, A₁₂ = 5, A₁₃ = 5
   • A₂₁ = 3, A₂₂ = −2, A₂₃ = −1
   • A₃₁ = 4, A₃₂ = −1, A₃₃ = −3
5. 5 · Write adj A and A⁻¹. adj A = [−5, 3, 4; 5, −2, −1; 5, −1, −3], so A⁻¹ = (1/5) adj A.
6. 6 · Find X = A⁻¹B. X = (1/5)[−10 + 24 − 4; 10 − 16 + 1; 10 − 8 + 3] = (1/5)[10; −5; 5] = [2; −1; 1].
7. 7 · Check and box the answer. 2 − 1 + 1 = 2, 4 + 1 + 3 = 8, 2 − 2 − 1 = −1. All hold. Answer: x = 2, y = −1, z = 1.

Scoring tip: Write |A| ≠ 0 before the cofactors, and check the answer in all three equations. Both are method marks.