# 15.3: Matrix Multiplication

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If $$\mathbf{A}$$ has dimensions $$m\times n$$ and $$\mathbf{B}$$ has dimensions $$n\times p$$, then the product $$\mathbf{AB}$$ is defined, and has dimensions $$m\times p$$.

The entry $$(ab)_{ij}$$ is obtained by multiplying row $$i$$ of $$\mathbf{A}$$ by column $$j$$ of $$\mathbf{B}$$, which is done by multiplying corresponding entries together and then adding the results:

Example $$\PageIndex{1}$$

Calculate the product

$\begin{pmatrix} 1 &-2 &4 \\ 5 &0 &3 \\ 0 & 1/2 &9 \end{pmatrix}\begin{pmatrix} 1 &0 \\ 5 &3 \\ -1 &0 \end{pmatrix} \nonumber$

Solution

We need to multiply a $$3\times 3$$ matrix by a $$3\times 2$$ matrix, so we expect a $$3\times 2$$ matrix as a result.

$\begin{pmatrix} 1 &-2 &4 \\ 5 &0 &3 \\ 0 & 1/2 &9 \end{pmatrix}\begin{pmatrix} 1 &0 \\ 5 &3 \\ -1 &0 \end{pmatrix}=\begin{pmatrix} a&b \\ c&d \\ e &f \end{pmatrix} \nonumber$

To calculate $$a$$, which is entry (1,1), we use row 1 of the matrix on the left and column 1 of the matrix on the right:

$\begin{pmatrix} {\color{red}1} &{\color{red}-2} &{\color{red}4} \\ 5 &0 &3 \\ 0 & 1/2 &9 \end{pmatrix}\begin{pmatrix} {\color{red}1} &0 \\ {\color{red}5} &3 \\ {\color{red}-1} &0 \end{pmatrix}=\begin{pmatrix} {\color{red}a}&b \\ c&d \\ e &f \end{pmatrix}\rightarrow a=1\times 1+(-2)\times 5+4\times (-1)=-13 \nonumber$

To calculate $$b$$, which is entry (1,2), we use row 1 of the matrix on the left and column 2 of the matrix on the right:

$\begin{pmatrix} {\color{red}1} &{\color{red}-2} &{\color{red}4} \\ 5 &0 &3 \\ 0 & 1/2 &9 \end{pmatrix}\begin{pmatrix} 1&{\color{red}0} \\ 5&{\color{red}3} \\ -1&{\color{red}0} \end{pmatrix}=\begin{pmatrix} a&{\color{red}b} \\ c&d \\ e &f \end{pmatrix}\rightarrow b=1\times 0+(-2)\times 3+4\times 0=-6 \nonumber$

To calculate $$c$$, which is entry (2,1), we use row 2 of the matrix on the left and column 1 of the matrix on the right:

$\begin{pmatrix} 1&-2&4\\ {\color{red}5} &{\color{red}0} &{\color{red}3} \\ 0 & 1/2 &9 \end{pmatrix}\begin{pmatrix} {\color{red}1} &0 \\ {\color{red}5} &3 \\ {\color{red}-1} &0 \end{pmatrix}=\begin{pmatrix} a&b \\ {\color{red}c}&d \\ e &f \end{pmatrix}\rightarrow c=5\times 1+0\times 5+3\times (-1)=2 \nonumber$

To calculate $$d$$, which is entry (2,2), we use row 2 of the matrix on the left and column 2 of the matrix on the right:

$\begin{pmatrix} 1&-2&4\\ {\color{red}5} &{\color{red}0} &{\color{red}3} \\ 0 & 1/2 &9 \end{pmatrix}\begin{pmatrix} 1&{\color{red}0} \\ 5&{\color{red}3} \\ -1&{\color{red}0} \end{pmatrix}=\begin{pmatrix} a&b \\ c&{\color{red}d} \\ e &f \end{pmatrix}\rightarrow d=5\times 0+0\times 3+3\times 0=0 \nonumber$

To calculate $$e$$, which is entry (3,1), we use row 3 of the matrix on the left and column 1 of the matrix on the right:

$\begin{pmatrix} 1&-2&4\\ 5&0&3 \\ {\color{red}0} &{\color{red}1/2} &{\color{red}9} \end{pmatrix}\begin{pmatrix} {\color{red}1} &0 \\ {\color{red}5} &3 \\ {\color{red}-1} &0 \end{pmatrix}=\begin{pmatrix} a&b \\ c&d \\ {\color{red}e} &f \end{pmatrix}\rightarrow e=0\times 1+1/2\times 5+9\times (-1)=-13/2 \nonumber$

To calculate $$f$$, which is entry (3,2), we use row 3 of the matrix on the left and column 2 of the matrix on the right:

$\begin{pmatrix} 1&-2&4\\ 5&0&3 \\ {\color{red}0} &{\color{red}1/2} &{\color{red}9} \end{pmatrix}\begin{pmatrix} 1&{\color{red}0} \\ 5&{\color{red}3} \\ -1&{\color{red}0} \end{pmatrix}=\begin{pmatrix} a&b \\ c&d \\ e&{\color{red}f} \end{pmatrix}\rightarrow f=0\times 0+1/2\times 3+9\times 0=3/2 \nonumber$

The result is:

$\displaystyle{\color{Maroon}\begin{pmatrix} 1 &-2 &4 \\ 5 &0 &3 \\ 0 & 1/2 &9 \end{pmatrix}\begin{pmatrix} 1 &0 \\ 5 &3 \\ -1 &0 \end{pmatrix}=\begin{pmatrix} -13&-6 \\ 2&0 \\ -13/2 &3/2 \end{pmatrix}} \nonumber$

Example $$\PageIndex{2}$$

Calculate

$\begin{pmatrix} 1 &-2 &4 \\ 5 &0 &3 \\ \end{pmatrix}\begin{pmatrix} 1 \\ 5 \\ -1 \end{pmatrix}\nonumber$

Solution

We are asked to multiply a $$2\times 3$$ matrix by a $$3\times 1$$ matrix (a column vector). The result will be a $$2\times 1$$ matrix (a vector).

$\begin{pmatrix} 1 &-2 &4 \\ 5 &0 &3 \\ \end{pmatrix}\begin{pmatrix} 1 \\ 5 \\ -1 \end{pmatrix}=\begin{pmatrix} a \\ b \end{pmatrix}\nonumber$

$a=1\times1+(-2)\times 5+ 4\times (-1)=-13\nonumber$

$b=5\times1+0\times 5+ 3\times (-1)=2\nonumber$

The solution is:

$\displaystyle{\color{Maroon}\begin{pmatrix} 1 &-2 &4 \\ 5 &0 &3 \\ \end{pmatrix}\begin{pmatrix} 1 \\ 5 \\ -1 \end{pmatrix}=\begin{pmatrix} -13 \\ 2 \end{pmatrix}}\nonumber$

Need help? The link below contains solved examples: Multiplying matrices of different shapes (three examples): http://tinyurl.com/kn8ysqq

## The Commutator

Matrix multiplication is not, in general, commutative. For example, we can perform

$\begin{pmatrix} 1 &-2 &4 \\ 5 &0 &3 \\ \end{pmatrix}\begin{pmatrix} 1 \\ 5 \\ -1 \end{pmatrix}=\begin{pmatrix} -13 \\ 2 \end{pmatrix} \nonumber$

but cannot perform

$\begin{pmatrix} 1 \\ 5 \\ -1 \end{pmatrix}\begin{pmatrix} 1 &-2 &4 \\ 5 &0 &3 \\ \end{pmatrix} \nonumber$

Even with square matrices, that can be multiplied both ways, multiplication is not commutative. In this case, it is useful to define the commutator, defined as:

$[\mathbf{A},\mathbf{B}]=\mathbf{A}\mathbf{B}-\mathbf{B}\mathbf{A} \nonumber$

Example $$\PageIndex{3}$$

Given $$\mathbf{A}=\begin{pmatrix} 3&1 \\ 2&0 \end{pmatrix}$$ and $$\mathbf{B}=\begin{pmatrix} 1&0 \\ -1&2 \end{pmatrix}$$

Calculate the commutator $$[\mathbf{A},\mathbf{B}]$$

Solution

$[\mathbf{A},\mathbf{B}]=\mathbf{A}\mathbf{B}-\mathbf{B}\mathbf{A}\nonumber$

$\mathbf{A}\mathbf{B}=\begin{pmatrix} 3&1 \\ 2&0 \end{pmatrix}\begin{pmatrix} 1&0 \\ -1&2 \end{pmatrix}=\begin{pmatrix} 3\times 1+1\times (-1)&3\times 0 +1\times 2 \\ 2\times 1+0\times (-1)&2\times 0+ 0\times 2 \end{pmatrix}=\begin{pmatrix} 2&2 \\ 2&0 \end{pmatrix}\nonumber$

$\mathbf{B}\mathbf{A}=\begin{pmatrix} 1&0 \\ -1&2 \end{pmatrix}\begin{pmatrix} 3&1 \\ 2&0 \end{pmatrix}=\begin{pmatrix} 1\times 3+0\times 2&1\times 1 +0\times 0 \\ -1\times 3+2\times 2&-1\times 1+2\times 0 \end{pmatrix}=\begin{pmatrix} 3&1 \\ 1&-1 \end{pmatrix}\nonumber$

$[\mathbf{A},\mathbf{B}]=\mathbf{A}\mathbf{B}-\mathbf{B}\mathbf{A}=\begin{pmatrix} 2&2 \\ 2&0 \end{pmatrix}-\begin{pmatrix} 3&1 \\ 1&-1 \end{pmatrix}=\begin{pmatrix} -1&1 \\ 1&1 \end{pmatrix}\nonumber$

$\displaystyle{\color{Maroon}[\mathbf{A},\mathbf{B}]=\begin{pmatrix} -1&1 \\ 1&1 \end{pmatrix}}\nonumber$

## Multiplication of a vector by a scalar

The multiplication of a vector $$\vec{v_1}$$ by a scalar $$n$$ produces another vector of the same dimensions that lies in the same direction as $$\vec{v_1}$$;

$n\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} nx \\ ny \end{pmatrix} \nonumber$

The scalar can stretch or compress the length of the vector, but cannot rotate it (figure [fig:vector_by_scalar]).

## Multiplication of a square matrix by a vector

The multiplication of a vector $$\vec{v_1}$$ by a square matrix produces another vector of the same dimensions of $$\vec{v_1}$$. For example, we can multiply a $$2\times 2$$ matrix and a 2-dimensional vector:

$\begin{pmatrix} a&b \\ c&d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} ax+by \\ cx+dy \end{pmatrix} \nonumber$

For example, consider the matrix

$\mathbf{A}=\begin{pmatrix} -2 &0 \\ 0 &1 \end{pmatrix} \nonumber$

The product

$\begin{pmatrix} -2&0 \\ 0&1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} \nonumber$

is

$\begin{pmatrix} -2x \\ y \end{pmatrix} \nonumber$

We see that $$2\times 2$$ matrices act as operators that transform one 2-dimensional vector into another 2-dimensional vector. This particular matrix keeps the value of $$y$$ constant and multiplies the value of $$x$$ by -2 (Figure $$\PageIndex{3}$$).

Notice that matrices are useful ways of representing operators that change the orientation and size of a vector. An important class of operators that are of particular interest to chemists are the so-called symmetry operators.

15.3: Matrix Multiplication is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Marcia Levitus via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.