6 The rules of matrix multiplication

It is worth noting that the process of multiplication can be continued to form products of more than two matrices.

Although two matrices may not commute (i.e. in general $AB\ne BA$ ) the associative law always holds i.e. for matrices which can be multiplied,

$A\left(BC\right)=\left(AB\right)C.$

The general principle is keep the left to right order , but within that limitation any two adjacent matrices can be multiplied.

It is important to note that it is not always possible to multiply together any two given matrices.

For example if $A=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]$ and $B=\left[\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill c\hfill \\ \hfill d\hfill & \hfill e\hfill & \hfill f\hfill \end{array}\right]$ then $AB=\left[\begin{array}{ccc}\hfill a+2d\hfill & \hfill b+2e\hfill & \hfill c+2f\hfill \\ \hfill 3a+4d\hfill & \hfill 3b+4e\hfill & \hfill 3c+4f\hfill \end{array}\right]$ .

However $BA=\left[\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill c\hfill \\ \hfill d\hfill & \hfill e\hfill & \hfill f\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]$ is not defined since each row of $B$ has three elements whereas each column of $A$ has two elements and we cannot multiply these elements in the manner described.

Task!

Given $A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill 5\hfill \\ \hfill 2\hfill & \hfill 4\hfill & \hfill 6\hfill \end{array}\right],B=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right],C=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 4\hfill \\ \hfill 2\hfill & \hfill 5\hfill \\ \hfill 3\hfill & \hfill 6\hfill \end{array}\right]$

State which of the products $AB,BA,AC,CA$ , $BC,CB,$ $\left(AB\right)C,$ $A\left(CB\right)$ is defined and state the size ( $n\times m$ ) of the product when defined.

Answer

We now list together some properties of matrix multiplication and compare them with corresponding properties for multiplication of numbers.

6.1 Application of matrices to networks

A network is a collection of points (nodes) some of which are connected together by lines (paths). The information contained in a network can be conveniently stored in the form of a matrix.

Example 5

Petrol is delivered to terminals $T_1$ and $T_2$ . They distribute the fuel to 3 storage depots ( $S_1,S_2,S_3$ ). The network diagram below shows what fraction of the fuel goes from each terminal to the three storage depots. In turn the 3 depots supply fuel to 4 petrol stations ( $P_1,P_2,P_3,P_4$ ) as shown in Figure 2:

Figure 2

Show how this situation may be described using matrices.

Solution

Denote the amount of fuel, in litres, flowing from $T_1$ by $t_1$ and from $T_2$ by $t_2$ and the quantity being received at $S_i$ by $s_i$  for $i=1,2,3$ . This situation is described in the following diagram:

From this diagram we see that

$\begin{array}{ccc}\hfill s_1\hfill & \hfill =\hfill & \hfill 0.4t_1+0.5t_2\hfill \\ \hfill s_2\hfill & \hfill =\hfill & \hfill 0.4t_1+0.2t_2\hfill \\ \hfill s_3\hfill & \hfill =\hfill & \hfill 0.2t_1+0.3t_2\hfill \\ \hfill \hfill \end{array}$ or, in matrix form: $\left[\begin{array}{c}\hfill s_1\hfill \\ \hfill s_2\hfill \\ \hfill s_3\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 0.4\hfill & \hfill 0.5\hfill \\ \hfill 0.4\hfill & \hfill 0.2\hfill \\ \hfill 0.2\hfill & \hfill 0.3\hfill \end{array}\right]\left[\begin{array}{c}\hfill t_1\hfill \\ \hfill t_2\hfill \\ \hfill \hfill \end{array}\right]$

In turn the 3 depots supply fuel to 4 petrol stations as shown in the next diagram:

If the petrol stations receive $p_1,p_2,p_3,p_4$ litres respectively then from the diagram we have:

$\begin{array}{ccc}\hfill p_1\hfill & \hfill =\hfill & \hfill 0.6s_1+0.2s_2\hfill \\ \hfill p_2\hfill & \hfill =\hfill & \hfill 0.2s_1+0.5s_2\hfill \\ \hfill p_3\hfill & \hfill =\hfill & \hfill 0.2s_1+0.2s_2+0.4s_3\hfill \\ \hfill p_4\hfill & \hfill =\hfill & \hfill 0.1s_2+0.6s_3\hfill \\ \hfill \hfill \end{array}$  or, in matrix form:  $\left[\begin{array}{c}\hfill p_1\hfill \\ \hfill p_2\hfill \\ \hfill p_3\hfill \\ \hfill p_4\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0.6\hfill & \hfill 0.2\hfill & \hfill 0\hfill \\ \hfill 0.2\hfill & \hfill 0.5\hfill & \hfill 0\hfill \\ \hfill 0.2\hfill & \hfill 0.2\hfill & \hfill 0.4\hfill \\ \hfill 0\hfill & \hfill 0.1\hfill & \hfill 0.6\hfill \end{array}\right]\left[\begin{array}{c}\hfill s_1\hfill \\ \hfill s_2\hfill \\ \hfill s_3\hfill \end{array}\right]$

Combining the equations, substituting expressions for $s_1,s_2,s_3$ in the equations for $p_1,p_2,p_3,p_4$

we get:

with similar results for $p_2,p_3$ and $p_4$ .

This is equivalent to combining the two networks. The results can be obtained more easily by multiplying the matrices: