• Home

Matrix Multiplication

You can only multiply two matrices if their dimensions are compatible , which means the number of columns in the first matrix is the same as the number of rows in the second matrix.

If $A=\left[a_{ij}\right]$ is an $m\times n$ matrix and $B=\left[b_{ij}\right]$ is an $n\times p$ matrix, the product $AB$ is an $m\times p$ matrix.

$AB=\left[c_{ij}\right]$ , where $c_{ij}=a_{i1}{b}_{1j}+a_{i2}{b}_{2j}+\mathrm{...}+a_{in}{b}_{nj}$ .

(The entry in the $i^{\text{th}}$ row and $j^{\text{th}}$ column is denoted by the double subscript notation $a_{ij}$ , $b_{ij}$ , and $c_{ij}$ . For instance, the entry $a_{23}$ is the entry in the second row and third column.)

The definition of matrix multiplication indicates a row-by-column multiplication, where the entries in the $i^{\text{th}}$ row of $A$ are multiplied by the corresponding entries in the $j^{\text{th}}$ column of $B$ and then adding the results.

Matrix multiplication is NOT commutative.   If neither $A$ nor $B$ is an identity matrix, $AB\ne BA$ .

Multiplying a Row by a Column

We'll start by showing you how to multiply a $1\times n$ matrix by an $n\times 1$ matrix. The first is just a single row, and the second is a single column. By the rule above, the product is a $1\times 1$ matrix; in other words, a single number.

First, let's name the entries in the row $r_1,r_2,\mathrm{...},r_n$ , and the entries in the column $c_1,c_2,\mathrm{...},c_n$ . Then the product of the row and the column is the $1\times 1$ matrix

$\left[r_1{c}_1+r_2{c}_2+\mathrm{...}+r_n{c}_n\right]$ .

We have to multiply a $1\times 3$ matrix by a $1\times 3$ matrix. The number of columns in the first is the same as the number of rows in the second, so they are compatible.

The product is:

$\begin{array}{l}\left[\left(1\right)\left(2\right)+\left(4\right)\left(-1\right)+\left(0\right)\left(5\right)\right]\\ =\left[2+\left(-4\right)+0\right]\\ =\left[-2\right]\end{array}$

Multiplying Larger Matrices

Now that you know how to multiply a row by a column, multiplying larger matrices is easy. For the entry in the $i^{\text{th}}$ row and the $j^{\text{th}}$ column of the product matrix, multiply each entry in the $i^{\text{th}}$ row of the first matrix by the corresponding entry in the $j^{\text{th}}$ column of the second matrix and adding the results.

Let's take the following problem, multiplying a $2\times 3$ matrix with a $3\times 2$ matrix, to get a $2\times 2$ matrix as the product. The entries of the product matrix are called $e_{ij}$ when they're in the $i^{\text{th}}$ row and $j^{\text{th}}$ column.

$\left[\begin{array}{rrr}\hfill 1& \hfill 0& \hfill 1\\ \hfill 0& \hfill 1& \hfill 2\end{array}\right]\cdot \left[\begin{array}{rr}\hfill 3& \hfill 5\\ \hfill -1& \hfill 0\\ \hfill 2& \hfill -1\end{array}\right]=\left[\begin{array}{rr}\hfill e_{11}& \hfill e_{12}\\ \hfill e_{21}& \hfill e_{22}\end{array}\right]$

To get $e_{11}$ , multiply Row $1$ of the first matrix by Column $1$ of the second.

$e_{11}=\left[\begin{array}{rrr}\hfill 1& \hfill 0& \hfill 1\end{array}\right]\cdot \left[\begin{array}{r}\hfill 3\\ \hfill -1\\ \hfill 2\end{array}\right]=1\left(3\right)+0\left(-1\right)+1\left(2\right)=5$

To get $e_{12}$ , multiply Row $1$ of the first matrix by Column $2$ of the second.

$e_{12}=\left[\begin{array}{rrr}\hfill 1& \hfill 0& \hfill 1\end{array}\right]\cdot \left[\begin{array}{r}\hfill 5\\ \hfill 0\\ \hfill -1\end{array}\right]=1\left(5\right)+0\left(0\right)+1\left(-1\right)=4$

To get $e_{21}$ , multiply Row $2$ of the first matrix by Column $1$ of the second.

$e_{21}=\left[\begin{array}{rrr}\hfill 0& \hfill 1& \hfill 2\end{array}\right]\cdot \left[\begin{array}{r}\hfill 3\\ \hfill -1\\ \hfill 2\end{array}\right]=0\left(3\right)+1\left(-1\right)+2\left(2\right)=3$

To get $e_{22}$ , multiply Row $2$ of the first matrix by Column $2$ of the second.

$e_{22}=\left[\begin{array}{rrr}\hfill 0& \hfill 1& \hfill 2\end{array}\right]\cdot \left[\begin{array}{r}\hfill 5\\ \hfill 0\\ \hfill 1\end{array}\right]=0\left(5\right)+1\left(0\right)+2\left(-1\right)=-2$

Writing the product matrix, we get:

$\left[\begin{array}{rr}\hfill e_{11}& \hfill e_{12}\\ \hfill e_{21}& \hfill e_{22}\end{array}\right]=\left[\begin{array}{rr}\hfill 5& \hfill 4\\ \hfill 3& \hfill -2\end{array}\right]$

Therefore, we have shown:

$\left[\begin{array}{rrr}\hfill 1& \hfill 0& \hfill 1\\ \hfill 0& \hfill 1& \hfill 2\end{array}\right]\cdot \left[\begin{array}{rr}\hfill 3& \hfill 5\\ \hfill -1& \hfill 0\\ \hfill 2& \hfill -1\end{array}\right]=\left[\begin{array}{rr}\hfill 5& \hfill 4\\ \hfill 3& \hfill -2\end{array}\right]$