WAEC 2018 · Paper 2 · Q7✱✱

(a)
Evaluate $\displaystyle\int_1^2 (2x^3 - 4x + 3)\,dx$ .
(b)
Given that $P^{-1} = \begin{pmatrix} -1 & 1 \\ 4 & -3 \end{pmatrix}$ , find the matrix $P$ .
Show the answer
$P = \begin{pmatrix} 3 & 1 \\ 4 & 1 \end{pmatrix}$

Worked solution (try it first)

Integrate term by term, adding 1 to each power and dividing by the new power: $\int (2x^3 - 4x + 3)\,dx = \frac{x^4}{2} - 2x^2 + 3x$ .
At $x = 2$ : $\frac{16}{2} - 8 + 6 = 6$ .
At $x = 1$ : $\frac12 - 2 + 3 = \frac32$ .
Top limit minus bottom limit: $6 - \frac32 = 4\frac12$ .

The inverse of $P^{-1}$ is $P$ , so find the inverse of $\begin{pmatrix} -1 & 1 \\ 4 & -3 \end{pmatrix}$ .
Its determinant is $(-1)(-3) - (1)(4) = 3 - 4 = -1$ .
Swap $-1$ and $-3$ , change the signs of $1$ and $4$ , and divide by $-1$ : $P = \frac{1}{-1}\begin{pmatrix} -3 & -1 \\ -4 & -1 \end{pmatrix}$
$= \begin{pmatrix} 3 & 1 \\ 4 & 1 \end{pmatrix}$ .
Check: $\begin{pmatrix} 3 & 1 \\ 4 & 1 \end{pmatrix}\begin{pmatrix} -1 & 1 \\ 4 & -3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ ✓.