WAEC 2023 · Paper 2 · Q10

(a)(i)
Calculate $\begin{vmatrix} 3 & 5 & -4 \\ 6 & -3 & -5 \\ -2 & 2 & 1 \end{vmatrix}$ .
(a)(ii)
Using your result in 10(a)(i), solve the simultaneous equations $3x + 5y - 4z = 1$ , $6x - 3y - 5z = -15$ , $-2x + 2y + z = 5$ . Enter $x, y, z$ .
Separate values with commas, e.g. 3, −2
(b)
Given that $x^2 + y^2 = 2pxy$ , where $p$ is a constant, find $\dfrac{dy}{dx}$ .
Show the answer
$\dfrac{py - x}{y - px}$

Worked solution (try it first)

(a)(i)

Expand along the top row: $3(-3 + 10) - 5(6 - 10) + (-4)(12 - 6) = 21 + 20 - 24$
$= 17$ .

(ii)

By Cramer's rule, replace each column by $(1, -15, 5)$ in turn: $\Delta_x = 17$ , $\Delta_y = 34$ and $\Delta_z = 51$ .
So $x = \dfrac{17}{17} = 1$ , $y = \dfrac{34}{17} = 2$ and $z = \dfrac{51}{17} = 3$ .
Check in the third equation: $-2 + 4 + 3 = 5$ ✓.

(b)

Differentiate both sides with respect to $x$ : $2x + 2y\dfrac{dy}{dx} = 2p\left(y + x\dfrac{dy}{dx}\right)$ .
Collect the $\dfrac{dy}{dx}$ terms: $2y\dfrac{dy}{dx} - 2px\dfrac{dy}{dx} = 2py - 2x$ .
Factorise and divide by 2: $(y - px)\dfrac{dy}{dx} = py - x$ , so $\dfrac{dy}{dx} = \dfrac{py - x}{y - px}$ .

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