# Multiplication of Two Matrices (Matrix Multiplication)

## Multiplication of Two Matrices (Matrix Multiplication)

Definition: Two matrices A and B are said to be conformable for multiplication. If the number of columns in A is equal to the number of rows in B. If A = (aij)m x n, B = (bjk) n x p are two matrices (which are conformable for multiplication). Then their product is defined to be matrices C = (cik) where $${{c}_{ik}}\,=\,\sum\limits_{j=1}^{n}{{{a}_{ij}}}{{b}_{jk}}$$. It is denoted by AB.

Example: Multiplication of two matrices $$A=\left[ \begin{matrix}{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\end{matrix} \right]$$, $$B=\left[ \begin{matrix}{{b}_{11}} & {{b}_{12}} \\{{b}_{21}} & {{b}_{22}} \\{{b}_{31}} & {{b}_{32}} \\\end{matrix} \right]$$.

Solution:

$$A=\left[ \begin{matrix}{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\end{matrix} \right]$$, $$B=\left[ \begin{matrix}{{b}_{11}} & {{b}_{12}} \\{{b}_{21}} & {{b}_{22}} \\{{b}_{31}} & {{b}_{32}} \\\end{matrix} \right]$$,

$$AB=\left[\begin{matrix}{{a}_{11}}{{b}_{11}}+{{a}_{12}}{{b}_{21}}+{{a}_{13}}{{b}_{31}}&{{a}_{11}}{{b}_{12}}+{{a}_{12}}{{b}_{22}}+{{a}_{13}}{{b}_{32}}\\{{a}_{21}}{{b}_{11}}+{{a}_{22}}{{b}_{21}}+{{a}_{23}}{{b}_{31}} &{{a}_{21}}{{b}_{12}}+{{a}_{22}}{{b}_{22}}+{{a}_{23}}{{b}_{32}} \\\end{matrix} \right]$$.

Properties of matrices multiplication:

1. If the product AB exists then it most necessary that product BA will also exist.
2. Matrices multiplication is not commutative. Even if AB and BA exist they need not be equal.
3. If AB = O then either A or B need not be equal to O.
4. If AB = AC then B not equal to C even if A ≠ 0.
5. Multiplication of matrices is associate. If A, B and C three matrices then (AB) C = A (BC).
6. Multiplication of the matrices is distribution over matrices addition i,e., if Conformability is assured for the matrices A, B and C then A (B + C) = AB + AC.
7. If A is not a matrices of types m x n then AIn = Im A = A.

Example: If $$A=\left[ \begin{matrix}1 & 1 & 3 \\5 & 2 & 6 \\-2 & -1 & -3 \\\end{matrix} \right]$$ then show that A³ = O.

Solution:

$$A=\left[ \begin{matrix}1 & 1 & 3 \\5 & 2 & 6 \\-2 & -1 & -3 \\\end{matrix} \right]$$,

$${{A}^{2}}=\left[ \begin{matrix}1 & 1 & 3 \\5 & 2 & 6 \\-2 & -1 & -3 \\\end{matrix} \right]\left[ \begin{matrix}1 & 1 & 3 \\5 & 2 & 6 \\-2 & -1 & -3 \\\end{matrix} \right]$$,

$$=\left[ \begin{matrix}1+5-6 & 1+2-3 & 3+6-9 \\5+10-12 & 5+4-6 & 15+12-18 \\-2-5+6 & -2-2+3 & -6-6+9 \\\end{matrix} \right]$$,

$$=\left[ \begin{matrix}0 & 0 & 0 \\3 & 3 & 9 \\-1 & -1 & -3 \\\end{matrix} \right]$$,

$${{A}^{3}}=\left[ \begin{matrix}0 & 0 & 0 \\3 & 3 & 9 \\-1 & -1 & -3 \\\end{matrix} \right]\left[ \begin{matrix}1 & 1 & 3 \\5 & 2 & 6 \\-2 & -1 & -3 \\\end{matrix} \right]$$,

$$=\left[ \begin{matrix}0+3-3 & 0+3-3 & 0+9-9 \\0+6-6 & 0+6-6 & 0+18-18 \\0-3+3 & 0-3+3 & 0-9+9 \\\end{matrix} \right]$$,

$$=\left[ \begin{matrix}0 & 0 & 0 \\0 & 0 & 0 \\0 & 0 & 0 \\\end{matrix} \right]$$.