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:
- If the product AB exists then it most necessary that product BA will also exist.
- Matrices multiplication is not commutative. Even if AB and BA exist they need not be equal.
- If AB = O then either A or B need not be equal to O.
- If AB = AC then B not equal to C even if A ≠ 0.
- Multiplication of matrices is associate. If A, B and C three matrices then (AB) C = A (BC).
- 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.
- 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:
\(A=\left[ \begin{matrix}1 & 1 & 3 \\5 & 2 & 6 \\-2 & -1 & -3 \\\end{matrix} \right]\),
\({{A}^{2}}=\left[ \begin{matrix}1 & 1 & 3 \\5 & 2 & 6 \\-2 & -1 & -3 \\\end{matrix} \right]\left[ \begin{matrix}1 & 1 & 3 \\5 & 2 & 6 \\-2 & -1 & -3 \\\end{matrix} \right]\),
\(=\left[ \begin{matrix}1+5-6 & 1+2-3 & 3+6-9 \\5+10-12 & 5+4-6 & 15+12-18 \\-2-5+6 & -2-2+3 & -6-6+9 \\\end{matrix} \right]\),
\(=\left[ \begin{matrix}0 & 0 & 0 \\3 & 3 & 9 \\-1 & -1 & -3 \\\end{matrix} \right]\),
\({{A}^{3}}=\left[ \begin{matrix}0 & 0 & 0 \\3 & 3 & 9 \\-1 & -1 & -3 \\\end{matrix} \right]\left[ \begin{matrix}1 & 1 & 3 \\5 & 2 & 6 \\-2 & -1 & -3 \\\end{matrix} \right]\),
\(=\left[ \begin{matrix}0+3-3 & 0+3-3 & 0+9-9 \\0+6-6 & 0+6-6 & 0+18-18 \\0-3+3 & 0-3+3 & 0-9+9 \\\end{matrix} \right]\),
\(=\left[ \begin{matrix}0 & 0 & 0 \\0 & 0 & 0 \\0 & 0 & 0 \\\end{matrix} \right]\).