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.

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