WAEC 2018 · Paper 2 · Q1

(a)
If $\begin{vmatrix} x - 3 & -4 & 3 \\ 5 & 2 & 2 \\ 2 & -4 & 6 - x \end{vmatrix} = -24$ , find the values of $x$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Expand along the first row, with the signs $+, -, +$ : $(x - 3)\begin{vmatrix} 2 & 2 \\ -4 & 6 - x \end{vmatrix} - (-4)\begin{vmatrix} 5 & 2 \\ 2 & 6 - x \end{vmatrix} + 3\begin{vmatrix} 5 & 2 \\ 2 & -4 \end{vmatrix}$ .
First minor: $2(6 - x) - 2(-4) = 12 - 2x + 8 = 20 - 2x$ .
Second minor: $5(6 - x) - 2 \times 2 = 30 - 5x - 4$
$= 26 - 5x$ .
Third minor: $5(-4) - 2 \times 2 = -24$ .
So $(x - 3)(20 - 2x) + 4(26 - 5x) + 3(-24) = -24$ .
Expand: $-2x^2 + 26x - 60 + 104 - 20x - 72 = -24$ , so $-2x^2 + 6x - 28 = -24$ .
Add 24 to both sides: $-2x^2 + 6x - 4 = 0$ .
Divide by $-2$ : $x^2 - 3x + 2 = 0$ .
Factorise: $(x - 1)(x - 2) = 0$ , so $x = 1$ or $x = 2$ .