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JAMB 2000 · UME · Q18

Find the values of tt for which the determinant of the matrix (t−400−1t+1134t−2)\begin{pmatrix} t - 4 & 0 & 0 \\ -1 & t + 1 & 1 \\ 3 & 4 & t - 2 \end{pmatrix} is zero.

Worked solution (try it first)
  1. The first row is (t−4,0,0)(t - 4, 0, 0), so expand along it: the determinant is (t−4)[(t+1)(t−2)−1×4](t - 4)\left[(t + 1)(t - 2) - 1 \times 4\right].
  2. Inside the bracket, t2−t−2−4=t2−t−6t^2 - t - 2 - 4 = t^2 - t - 6, which factorises as (t−3)(t+2)(t - 3)(t + 2).
  3. So the determinant is (t−4)(t−3)(t+2)(t - 4)(t - 3)(t + 2), which is zero when t=4t = 4, 3 or −2-2, option D.

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