Q.If $A=\begin{pmatrix}1&2&3\\0&1&4\end{pmatrix}$ (order $2\times3$) and $B=\begin{pmatrix}2&0\\1&3\\0&2\end{pmatrix}$ (order $3\times2$), find $AB$.

Since the number of columns of $A$ (3) equals the number of rows of $B$ (3), the product $AB$ is defined and has order $2\times2$; each element is found by the row-by-column rule.

Conformability check

$A$ is $2\times3$ and $B$ is $3\times2$; since $A$'s columns (3) equal $B$'s rows (3), $AB$ is defined and will be of order $2\times2$.

Row 1 of $A$ with each column of $B$

$(1,1)$ entry: row 1 of $A$ $(1,2,3)$ with column 1 of $B$ $(2,1,0)$:

$(1,2)$ entry: row 1 of $A$ with column 2 of $B$ $(0,3,2)$:

Row 2 of $A$ with each column of $B$

$(2,1)$ entry: row 2 of $A$ $(0,1,4)$ with column 1 of $B$ $(2,1,0)$:

$(2,2)$ entry: row 2 of $A$ with column 2 of $B$ $(0,3,2)$:

$AB=\begin{pmatrix}4&12\\1&11\end{pmatrix}$ …

Multiplying entries that are not in the correct row-column combination (e.g. pairing row 1 of A with row 1 of B instead of a COLUMN of B), or forgettin …