WAEC 2009 · Paper 2 · Q12✱✱

(a)(i)
Two linear transformations are defined by $P : (x, y) \to (-2x + 3y,\ 5x - 4y)$ and $T : (x, y) \to (-3x - 4y,\ 6x - 5y)$ . Find the inverse of $T$ ;
Separate values with commas, e.g. 3, −2
(a)(ii)
the image of $(1, 3)$ under the transformation $P$ .
Separate values with commas, e.g. 3, −2
(b)
Find the volume, in terms of $\pi$ , of the solid formed when the area enclosed by the lines $y = 1$ , $2y = x + 1$ and $x = 0$ is rotated through two right angles about the $y$ -axis.

Worked solution (try it first)

The matrix of $T$ is $\begin{pmatrix} -3 & -4 \\ 6 & -5 \end{pmatrix}$ , with determinant $(-3)(-5) - (-4)(6) = 15 + 24 = 39$ .
Swap the leading diagonal, change the signs of the others and divide by 39: $T^{-1} = \frac{1}{39}\begin{pmatrix} -5 & 4 \\ -6 & -3 \end{pmatrix}$ .
So $T^{-1} : (x, y) \to \left(\dfrac{-5x + 4y}{39},\ \dfrac{-6x - 3y}{39}\right) = \left(\dfrac{-5x + 4y}{39},\ \dfrac{-2x - y}{13}\right)$ .

The matrix of $P$ is $\begin{pmatrix} -2 & 3 \\ 5 & -4 \end{pmatrix}$ .
Multiply by $\begin{pmatrix} 1 \\ 3 \end{pmatrix}$ : $\begin{pmatrix} -2 + 9 \\ 5 - 12 \end{pmatrix} = \begin{pmatrix} 7 \\ -7 \end{pmatrix}$ .
The image is $(7, -7)$ .

The region is the triangle with corners $(0, \frac12)$ , $(0, 1)$ and $(1, 1)$ .
Along it, $x = 2y - 1$ for $\frac12 \le y \le 1$ .
Two right angles is half a turn, so the volume is half of $\pi\int x^2\,dy$ : $V = \dfrac{\pi}{2}\displaystyle\int_{1/2}^{1} (2y - 1)^2\,dy$ .
Expand and integrate: $\dfrac{\pi}{2}\left[\frac43y^3 - 2y^2 + y\right]_{1/2}^{1} = \dfrac{\pi}{2}\left(\frac13 - \frac16\right)$ .
So $V = \dfrac{\pi}{2} \times \dfrac16$
$= \dfrac{\pi}{12}$ cubic units.