Find the values of x for which |x, 2, 1; 2, x, 0; 1, 1, 1| = 0.

1. 1 · Write what is given and what is to be found. Given: the 3 × 3 determinant equals 0. To find: all values of x.
2. 2 · Write the rule. Expand along a row: Δ = a₁₁M₁₁ − a₁₂M₁₂ + a₁₃M₁₃ (signs + − +).
3. 3 · Expand along row 1.
   • x(x × 1 − 0 × 1) = x²
   • − 2(2 × 1 − 0 × 1) = −4
   • + 1(2 × 1 − x × 1) = 2 − x
4. 4 · Form the equation. x² − 4 + 2 − x = 0, so x² − x − 2 = 0.
5. 5 · Solve it. (x − 2)(x + 1) = 0, so x = 2 or x = −1.
6. 6 · Check and box the answer. Put x = 2: 2(2) − 2(2) + 1(2 − 2) = 0. Answer: x = 2 or x = −1.

Scoring tip: Write the + − + signs above the row before you expand. Most lost marks come from a wrong middle sign.