WAEC 2023 · Paper 1 · Q24

Matrices & linear transformations
Polynomials & quadratic roots

Given that $P = \begin{pmatrix} x & 4 \\ 3 & 7 \end{pmatrix}$ , $Q = \begin{pmatrix} x & 3 \\ 1 & 2x \end{pmatrix}$ and the determinant of $Q$ is three more than that of $P$ , find the values of $x$ .

Worked solution (try it first)

$|P| = 7x - 12$ and $|Q| = 2x^2 - 3$ .
$|Q|$ is three more than $|P|$ : $2x^2 - 3 = 7x - 12 + 3$ , which gives $2x^2 - 7x + 6 = 0$ .
Factorise: $(2x - 3)(x - 2) = 0$ .
So $x = 2$ or $x = \frac{3}{2}$ , option D.

Report a problem with this question