Multiplication of Two Matrices

Multiplication of Two Matrices

Two matrices $A_{m\times n}$ and $B_{n\times p}$ are conformable for the product $AB$ exactly when the number of columns of $A$ equals the number of rows of $B$. The product is then $C=AB$ of order $m\times p$, with entries

— that is, $C_{ij}$ is "row $i$ of $A$ dotted with column $j$ of $B$".

Worked Examples

Example 1. $A=[a_{11}\ a_{12}\ a_{13}]_{1\times3}$, $B=\begin{bmatrix}b_{11}\\b_{21}\\b_{31}\end{bmatrix}_{3\times1}$: since columns of $A$ = rows of $B$ = 3, $AB$ is defined, order $1\times1$: $AB=[a_{11}b_{11}+a_{12}b_{21}+a_{13}b_{31}]$.

Example 2. $A=[1\ \ 3\ \ 2]_{1\times3}$, $B=\begin{bmatrix}3\\2\\1\end{bmatrix}_{3\times1}$.

Step 1: $AB=[1\times3+3\times2+2\times1]=[3+6+2]=[11]_{1\times1}$.

Step 2: $BA$ is also defined (columns of $B$=1=rows of $A$), order $3\times3$: $BA=\begin{bmatrix}3\\2\\1\end{bmatrix}[1\ \ 3\ \ 2]=\begin{bmatrix}3&9&6\\2&6&4\\1&3&2\end{bmatrix}$.

Example 3. $A=\begin{bmatrix}1&-2\\2&-1\\3&0\end{bmatrix}_{3\times2}$, $B=\begin{bmatrix}1&-2\\-1&2\end{bmatrix}_{2\times2}$.

Step 1: $A$ is $3\times2$, $B$ is $2\times2$, so $AB$ (order $3\times2$) is defined but $BA$ is NOT (columns of $B$=2 ≠ rows of $A$=3).

Step 2: $AB=\begin{bmatrix}1(1)+(-2)(-1)&1(-2)+(-2)(2)\\2(1)+(-1)(-1)&2(-2)+(-1)(2)\\3(1)+0(-1)&3(-2)+0(2)\end{bmatrix}=\begin{bmatrix}3&-6\\3&-6\\3&-6\end{bmatrix}$.

Example 4. $A=\begin{bmatrix}3&-2&1\\2&5&4\end{bmatrix}_{2\times3}$, $B=\begin{bmatrix}-3&2\\3&-2\end{bmatrix}_{2\times2}$: since columns of $A$(3) ≠ rows of $B$(2), $AB$ is not defined; but columns of $B$(2)=rows of $A$(2), so $BA$ (order $2\times3$) is defined and can be computed by the row-times-column rule.

Example 5. $A=\begin{bmatrix}4&5\\3&2\end{bmatrix}$, $B=\begin{bmatrix}-1&3\\4&-2\end{bmatrix}$ — both $2\times2$, so both $AB$ and $BA$ exist and are $2\times2$. …