3.4 Multiplication of Matrices

3.4 Multiplication of Matrices

A matrix $A$ is said to be conformable for multiplication with a matrix $B$ if the number of columns of $A$ is equal to the number of rows of $B$.

Thus, if $A$ and $B$ are given two matrices of order $m$-by-$n$ and $n$-by-$p$ respectively, then the product of matrices $A$ and $B$ is denoted by $AB$ and its order will be $m$-by-$p$.

The order of $AB$ is $m$-by-$p$ = number of rows of $A$ by number of columns of $B$.

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Remember!
Number of rows of $A$ $\rightarrow$ $A$ $m$-by-$n$ must be same $\rightarrow$ $B$ $n$-by-$p$ $\rightarrow$ Number of columns of $B$
Product $AB$ is of order $m$-by-$p$

Let us explain the concept of multiplication of matrices with the help of an example from our daily life.

Talmeez purchased $7$ books and $6$ pencils at the rate of Rs. $20$ and Rs. $3$ per article respectively from shop No. $1$ and he purchased same number of books and pencils at the rate of Rs. $22$ and Rs. $4$ per article respectively from Shop No. $2$. Majid purchased $5$ books and $8$ pencils from Shop No. $1$ at the same rate Talmeez purchased. Majid purchased same number of books and pencils from Shop No. $2$ at the same rate Talmeez purchased.

We show their purchase in the form of tables.

Table 1

Table 2

Amount paid by Talmeez at Shop No. $1$.

$7 \times 20 + 6 \times 3 = 158$ rupees ...(i)

Amount paid by Talmeez at Shop No. $2$

$7 \times 22 + 6 \times 4 = 178$ rupees ...(ii)

Amount paid by Majid at shop No. $1$

$5 \times 20 + 8 \times 3 = 124$ rupees ...(iii)

Amount paid by Majid at shop No. $2$

$5 \times 22 + 8 \times 4 = 142$ rupees ...(iv)

We show entries of the two tables in the form of matrices.

Table 1

$\begin{bmatrix} 7 & 6 \\ 5 & 8 \end{bmatrix}$

Table 2

$\begin{bmatrix} 20 & 22 \\ 3 & 4 \end{bmatrix}$

Calculations given above is shown in the form of matrix.

$\begin{bmatrix} 7 \times 20 + 6 \times 3 & 7 \times 22 + 6 \times 4 \\ 5 \times 20 + 8 \times 3 & 5 \times 22 + 8 \times 4 \end{bmatrix}$ see(i),(ii),(iii)and (iv)

$= \begin{bmatrix} 158 & 178 \\ 124 & 142 \end{bmatrix}$

For example, $A = \begin{bmatrix} 9 & 7 \\ 4 & 3 \end{bmatrix}, B = \begin{bmatrix} 2 & 6 \\ 5 & 8 \end{bmatrix}$

Now, $AB = \begin{bmatrix} 9 & 7 \\ 4 & 3 \end{bmatrix} \begin{bmatrix} 2 & 6 \\ 5 & 8 \end{bmatrix}$

Example 5

If $A = \begin{bmatrix} 2 & 0 \\ 5 & -3 \end{bmatrix}, B = \begin{bmatrix} -2 \\ 1 \end{bmatrix}$, then find:
(i) $AB$, if possible. (ii) $BA$, if possible.

Solution (i) Number of columns in matrix $A = 2$

Number of rows in matrix $B = 2$

Thus, $AB$ is possible

Now $AB = \begin{bmatrix} 2 & 0 \\ 5 & -3 \end{bmatrix} \begin{bmatrix} -2 \\ 1 \end{bmatrix}$

$= \begin{bmatrix} 2 \times (-2) + 0 \times 1 \\ 5 \times (-2) + (-3) \times 1 \end{bmatrix}$

$= \begin{bmatrix} -4 + 0 \\ -10 - 3 \end{bmatrix} = \begin{bmatrix} -4 \\ -13 \end{bmatrix}$

(ii) $BA = \begin{bmatrix} -2 \\ 1 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 5 & -3 \end{bmatrix}$

Number of columns in matrix $B = 1$

Number of rows in matrix $A = 2$

Thus, $BA$ is not possible

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Skilled Practice!
If $A = [5 \quad 3]$ and $B = \begin{bmatrix} -2 \\ 1 \end{bmatrix}$, then find $AB$ and its order.