JAMB 2004 · UME · Q23

If $P = \begin{pmatrix} 1 & 0 & -1 \\ 3 & 4 & 5 \\ -1 & 0 & 1 \end{pmatrix}$ , then $|P|$ is

Worked solution (try it first)

Expand along the first row, with signs $+ \; - \; +$ .
The first term is $1 \times (4 \times 1 - 5 \times 0) = 4$ .
The second term is 0, because the entry is 0.
The third term is $-1 \times (3 \times 0 - 4 \times (-1)) = -1 \times 4$
$= -4$ .
So $|P| = 4 - 4 = 0$ , option B.
(Row 3 is row 1 times $-1$ , and that always makes the determinant 0.)