# Multiplication of Matrices – Examples and Definition

Here you will learn multiplication of matrices with definition and examples.

Let’s begin –

## Multiplication of Matrices

Definition : Two matrices A and B are conformable for the product AB if the number of columns in A is same as the number of row in B.

Let matrix A is of order $$m\times n$$ then m is the number of rows and n is the number of coumns in A

and matrix B is of order $$n\times p$$ then n is the number of rows and p is the number of columns in B.

Thus, if A = $$[a_{ij}]_{m\times n}$$ and B =  $$[b_{ij}]_{n\times p}$$ are two matrices of order $$m\times n$$ and $$n\times p$$ respectively, then their product AB is conformable and of order $$m\times p$$ and is defined as

$$(AB)_{ij}$$ = ( $$i^{th}$$ row of A) (( $$j^{th}$$ column of B)  for all i = 1, 2, ….., m and j = 1, 2, ……. , p.

$$\implies$$  $$(AB)_{ij}$$ = $$\begin{bmatrix} a_{i1} & a_{i2} & …… & a_{in} \end{bmatrix}$$ $$\begin{bmatrix} b_{1j} \\ b_{2j} \\ . \\ . \\ . \\ b_{nj} \end{bmatrix}$$

Note : If A and B are two matrices such that AB exists, then BA may or may not exist.

## Example :

If A = $$\begin{bmatrix} 2 & 1 & 3 \\ 3 & -2 & 1 \\ -1 & 0 & 1 \end{bmatrix}$$ and B = $$\begin{bmatrix} 1 & -2 \\ 2 & 1 \\ 4 & -3 \end{bmatrix}$$, then A is a $$3\times 3$$ matrix and B is a $$3\times 2$$ matrix. Therefore, A and b are conformable for the product AB and it is of order $$3\times 2$$ such that

$$(AB)_{11}$$ = (first row of A)(first column of B)

$$\implies$$ $$(AB)_{11}$$  = $$\begin{bmatrix} 2 & 1 & 3 \end{bmatrix}$$ $$\begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}$$

= 2*1 + 1*2 + 3*4 = 16

$$(AB)_{12}$$ = (first row of A)(second column of B)

$$\implies$$ $$(AB)_{12}$$  = $$\begin{bmatrix} 2 & 1 & 3 \end{bmatrix}$$ $$\begin{bmatrix} -2 \\ 1 \\ 3 \end{bmatrix}$$

= 2*(-2) + 1*1 + 3*(-3) = -12

$$(AB)_{21}$$ = (second row of A)(first column of B)

$$\implies$$ $$(AB)_{21}$$  = $$\begin{bmatrix} 3 & -2 & 1 \end{bmatrix}$$ $$\begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}$$

= 3*1 + (-2)*2 + 1*(4) = 3

Similarly, we obtain

$$(AB)_{22}$$ = -11 , $$(AB)_{31}$$ = 3 and $$(AB)_{32}$$ = -1

$$\therefore$$ AB = $$\begin{bmatrix} 16 & -12 \\ 3 & -11 \\ 3 & -1 \end{bmatrix}$$

Note : In this case BA does not exist, because the number of columns in B is not same as the number of rows in A.