NECO 2022 · Paper 1 · Q18

Given that $A = \begin{pmatrix} 2 & -\frac{1}{3} \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 2 \\ 3 & 6 \end{pmatrix}$ , calculate $AB$ .

Worked solution (try it first)

Row 1 of $A$ times the columns of $B$ : $2(1) + \left(-\frac{1}{3}\right)(3) = 1$ and $2(2) + \left(-\frac{1}{3}\right)(6) = 2$ .
Row 2 of $A$ times the columns of $B$ : $3(1) + 4(3) = 15$ and $3(2) + 4(6) = 30$ .
So $AB = \begin{pmatrix} 1 & 2 \\ 15 & 30 \end{pmatrix}$ , option A.