Find the inverse of A = [3, 0, 1; 2, 1, 0; 1, 1, 1] and verify that AA⁻¹ = I.

1. 1 · Write what is given and what is to be found. Given: A, a 3 × 3 matrix. To find: A⁻¹, and check AA⁻¹ = I.
2. 2 · Write the formula. A⁻¹ = (1/|A|) adj A, valid when |A| ≠ 0.
3. 3 · Find |A|. Along row 1: 3(1 − 0) − 0 + 1(2 − 1) = 3 + 1 = 4 ≠ 0, so A⁻¹ exists.
4. 4 · Find the cofactors.
   • A₁₁ = 1, A₁₂ = −2, A₁₃ = 1
   • A₂₁ = 1, A₂₂ = 2, A₂₃ = −3
   • A₃₁ = −1, A₃₂ = 2, A₃₃ = 3
5. 5 · Write adj A and A⁻¹. adj A = [1, 1, −1; −2, 2, 2; 1, −3, 3], so A⁻¹ = (1/4)[1, 1, −1; −2, 2, 2; 1, −3, 3].
6. 6 · Check and box the answer. A × adj A: row 1 gives 3 + 0 + 1 = 4, 3 + 0 − 3 = 0 and −3 + 0 + 3 = 0; the other rows give 0, 4, 0 and 0, 0, 4. So A × adj A = 4I and AA⁻¹ = I.

Scoring tip: Write '|A| ≠ 0, so A⁻¹ exists' before finding the cofactors. It is a method mark.