Matrix Multiplication

WHAT NEEDS TO HOLD

Matrix multiplication is quite different from what you might expect. You do NOT simply multiply corresponding entries multiply corresponding entries in the two matrices... it's a bit more complicated. Because of this there's an important condition that must hold: when finding the product of $AB$, the number of columns in $A$, must equal the number of rows in $B$ (the reason for this will make more sense once you learn the technique below). The resulting matrix will have the same number of rows as $A$ and the same number of columns as $B$ (i.e., if you multiply an $m \times n$ times an $n \times p$ matrix you will get a matrix of size $m \times p$.

Example: Let's say we have 3 different matrices: $A$, which is $3 \times 5$, $B$ which is $2 \times 3$ and $C$ which is $ 4 \times 3 $. Which of the following are defined: $AB , BA, BC, CA $? If it is defined what size will it be?

Solution:

$AB$ means we multiply a $3 \times 5$ by a $2 \times 3$. Since the number of rows in $B$ is 2 while the number of columns in $A$ is 5 the multiplication is NOT possible.

$BA$ means we multiply a $2 \times 3$ by a$3 \times 5$. Since the number of rows in $A$ is 3 while the number of columns in $B$ is 3 the multiplication is possible. The size of $BA$ is $2 \times 5$.

$BC$ means we multiply a $2 \times 3$ by a $4 \times 3$. Since the number of rows in $C$ is 4 while the number of columns in $B$ is 3 the multiplication is NOT possible.

$CA$ means we multiply a $4 \times 3$ by a $3 \times 5$. Since the number of rows in $A$ is 3 while the number of columns in $C$ is 3 the multiplication is possible. The size of $CA$ is $4 \times 5$.

HOW TO MULTIPLY MATRICES

As already mentioned , the technique might not be so intuitive. Here's what to do: If you have two matrices $A$ and $B$ and you want to obtain a new matrix $C = AB$, to get the entry $c_{ij}$, you multiply each entry in row $i$ of matrix $A$ by the corresponding entry in column $j$ of matrix $B$ and add them up.
e.g. if we multiply $2 \times 2$ matrices and we want to find entry $c_{21}$, we multiply entries in the 2nd of
$A$ by those in the 1st column of $B$ and add up the results.

$ \text{i.e.}
\quad c_{21} = a_{21}b_{11} + a_{22}b_{21}
$

e.g. to get $c_{21}$ we have: \[
\begin{bmatrix}
a_{11} & a_{12} \\
\colorbox{yellow}{$a_{21}$} & \colorbox{yellow}{$a_{22}$}\\
\end{bmatrix}
\begin{bmatrix}
\colorbox{yellow}{$b_{11}$} & b_{12} \\
\colorbox{yellow}{$b_{21}$}& b_{22} \\
\end{bmatrix}
\]

To find $c_{11}$: $
\begin{bmatrix}
\colorbox{yellow}{$a_{11}$} & \colorbox{yellow}{$a_{12}$} \\
a_{21} & a_{22}\\
\end{bmatrix}
\begin{bmatrix}
\colorbox{yellow}{$b_{11}$} & b_{12} \\
\colorbox{yellow}{$b_{21}$}& b_{22} \\
\end{bmatrix}
$

To find $c_{12}$: $
\begin{bmatrix}
\colorbox{yellow}{$a_{11}$} & \colorbox{yellow}{$a_{12}$} \\
a_{21} & a_{22}\\
\end{bmatrix}
\begin{bmatrix}
b_{11} & \colorbox{yellow}{$b_{12}$} \\
b_{21}& \colorbox{yellow}{$b_{22}$} \\
\end{bmatrix}
$

To find $c_{22}$: $
\begin{bmatrix}
a_{11} & a_{12} \\
\colorbox{yellow}{$a_{21}$} & \colorbox{yellow}{$a_{22}$}\\
\end{bmatrix}
\begin{bmatrix}
b_{11} & \colorbox{yellow}{$b_{12}$} \\
b_{21}& \colorbox{yellow}{$b_{22}$} \\
\end{bmatrix}
$

Example : If $ A = \begin{bmatrix}
3 & 5 \\
-1 & 0 \\
\end{bmatrix}
$ and $
B = \begin{bmatrix}
4 & -6 \\
2 & 1 \\
\end{bmatrix}
$ find $AB$.

Solution:
\[
\begin{equation}
\begin{split}
AB & = \begin{bmatrix}
(3)(4) + (5)(2) & (3)(-6) + (5)(1) \\
(-1)(4) + (0)(2) & (-1)(-6) + (0)(1) \\
\end{bmatrix} \\
& = \begin{bmatrix}
22 & -13 \\
-4 & 6 \\
\end{bmatrix} \\
\end{split}
\end{equation}
\]

Example: If $ A = \begin{bmatrix}
5 & -1 \\
6 & 2 \\
0 & 1 \\
\end{bmatrix}
$ and $ B = \begin{bmatrix}
5 & 1 \\
-7 & 2 \\
\end{bmatrix}
$ find $AB$.

Solution: Since $A$ is $3 \times 2$ and $B$ is $2 \times 2$, we are able to do multiplication and
$AB$ will be of the size $3 \times 2$.

\[
\begin{equation}
\begin{split}
AB & = \begin{bmatrix}
(5)(5) + (-1)(-7) & (5)(1) + (-1)(2) \\
(6)(5) + (2)(-7) & (6)(1) + (2)(2) \\
(0)(5) + (1)(-7) & (0)(1) + (1)(2) \\
\end{bmatrix} \\
& = \begin{bmatrix}
32 & 3 \\
16 & 10 \\
-7 & 2 \\
\end{bmatrix} \\
\end{split}
\end{equation}

\]

Example 1: 2x2

Example 2: 3x3

Example 3: 3x3 by 3x1

Example 4: 4x3 by 3x2