## Algebra and Trigonometry 10th Edition

$AB$ is an $3\times2$ matrix. $AB=\begin{bmatrix} -2 & 17 \\ 15 & 12 \\ 18 & 46 \end{bmatrix}$
$A$ is an $3\times3$ matrix and $B$ is an $3\times2$ matrix. The number of columns of $A$ is equal to the number of rows of $B$. So, it is possible to find $AB$, where $AB$ is a $3\times2$ matrix. $AB=\begin{bmatrix} 0 & -1 & 2 \\ 6 & 0 & 3 \\ 7 & -1 & 8 \end{bmatrix}·\begin{bmatrix} 2 & -1 \\ 4 & -5 \\ 1 & 6 \end{bmatrix}=\begin{bmatrix} 0(2)-1(4)+2(1) & 0(-1)-1(-5)+2(6) \\ 6(2)+0(4)+3(1) & 6(-1)+0(-5)+3(6) \\ 7(2)-1(4)+8(1) & 7(-1)-1(-5)+8(6) \end{bmatrix}=\begin{bmatrix} -2 & 17 \\ 15 & 12 \\ 18 & 46 \end{bmatrix}$