JAMB 1998 · UME · Q23

The determinant of the matrix $\begin{pmatrix} x & 1 & 0 \\ 1 - x & 2 & 3 \\ 1 & 1 + x & 4 \end{pmatrix}$ in terms of $x$ is

Worked solution (try it first)

Expand along the first row, with signs $+ \; - \; +$ .
The first term is $x(2 \times 4 - 3(1 + x)) = x(5 - 3x)$ , which is $5x - 3x^2$ .
The second term is $-1 \times (4(1 - x) - 3 \times 1) = -(1 - 4x)$ , which is $4x - 1$ .
The third term is 0, because the entry is 0.
Add them: $5x - 3x^2 + 4x - 1 = -3x^2 + 9x - 1$ , option B.