WAEC 2020 · Paper 2 · Q6

(a)
Given that $(y + 2)$ , $(y + 3)$ and $(2y^2 + 1)$ are consecutive terms of an Arithmetic Progression (A.P.), find the possible values of $y$ .
Separate values with commas, e.g. 3, −2
(b)
Given that $X = \begin{pmatrix} 2 & 4 \\ 8 & 6 \end{pmatrix}$ , $Y = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $XY = \begin{pmatrix} 12 & 16 \\ 8 & 10 \end{pmatrix}$ , find: (i) matrix $Y$ ; (ii) the determinant of $Y$ .

Worked solution (try it first)

Consecutive terms of an A.P. have equal gaps: $(y + 3) - (y + 2) = (2y^2 + 1) - (y + 3)$ .
Simplify each side: $1 = 2y^2 - y - 2$ .
Move everything to one side: $2y^2 - y - 3 = 0$ .
Factorise: $(2y - 3)(y + 1) = 0$ , so $y = \frac32$ or $y = -1$ .
(Check: $y = -1$ gives $1, 2, 3$ and $y = \frac32$ gives $3\frac12, 4\frac12, 5\frac12$ .)

Multiply row by column: $XY = \begin{pmatrix} 2a + 4c & 2b + 4d \\ 8a + 6c & 8b + 6d \end{pmatrix}$
$= \begin{pmatrix} 12 & 16 \\ 8 & 10 \end{pmatrix}$ .
The first column gives $2a + 4c = 12$ and $8a + 6c = 8$ .
Multiply the first by 4: $8a + 16c = 48$ .
Take away the second: $10c = 40$ , so $c = 4$ , and $2a = 12 - 16 = -4$ , so $a = -2$ .
The second column gives $2b + 4d = 16$ and $8b + 6d = 10$ .
Multiply the first by 4: $8b + 16d = 64$ .
Take away the second: $10d = 54$ , so $d = \frac{27}{5}$ , and $2b = 16 - \frac{108}{5} = -\frac{28}{5}$ , so $b = -\frac{14}{5}$ .
So $Y = \begin{pmatrix} -2 & -2\frac45 \\ 4 & 5\frac25 \end{pmatrix}$ .

$|Y| = ad - bc$
$= (-2) \times \frac{27}{5} - \left(-\frac{14}{5}\right) \times 4$
$= -\frac{54}{5} + \frac{56}{5}$
$= \frac25$ .