2.2 (b) Multiplication of Matrices:

Maths > Matrices and Determinants > 2.0 Algebra of Matrices

  Matrices and Determinants
    1.0 Introduction
    2.0 Algebra of Matrices
    3.0 Special Matrices
    4.0 Determinant of a square matrix
    5.0 Adjoint of a square Matrix
    6.0 Inverse of a Matrix
    7.0 Types of Equations Homogenous & Non-Homogenous
    8.0 Cramer's rule
    9.0 Types of Linear Equations
2.2 (b) Multiplication of Matrices:
Any two matrices can be multiplied only when the number of columns in the first matrix is equal to the number of rows in the second matrix.

Example:

If $A$= $\left[ {{a_{ij}}} \right]$ be an $m$ x $n$ matrix and $B$=$\left[ {{b_{ij}}} \right]$ be an $n$ x $p$ matrix,then the product $AB$ of these matrices is $m$ x $p$ matrix
$C$ = $\left[ {{c_{ij}}} \right]$

where $${c_{ij}} = \sum\limits_{k = 1}^n {{a_{ik}}} {b_{kj}}$$$$={a_{i1}}{b_{1j}} + {a_{i2}}{b_{2j}} + ....... + {a_{in}}{b_{nj}}$$

Properties of Multiplication

The Matrix $AB$ is the matrix $B$ pre-multiplied by $A$ and the matrix $BA$ is the matrix $B$ post-multiplied by $A$
  • Matrix multiplication may or may not be commutative i.e., $AB$ may or may not be equal to $BA$
(a)If $AB = BA$,then matrices $A$ and $B$ are called Commutative matrices

(b)If $AB $ \ne $ BA$,then matrices $A$ and $B$ are called Anti-Commutative matrices
  • Matrix multiplication is Associative i.e.,$(AB)C = A(BC)$
  • Matrix multiplication is Distributive over Matrix Addition i.e., $(A + B)C = AC + BC$
  • Cancellation Laws not necessarily hold in case of matrix multiplication i.e., if $AB = AC$ need not mean $B = C$,even if $A \ne 0$
  • $AB = 0$ i.e.,$AB$ is a Null Matrix,does not necessarily imply that either $A$ or $B$ is a null matrix but either $\left| A \right| = 0$ or $\left| B \right| = 0$ or both $\left| A \right| = \left| B \right| = 0$.In addition, if none of them are null matrices,then $\left| A \right| = \left| B \right| = 0$


Question 3.

If $A = \left[ {\begin{array}{c} 1&2 \\ { - 2}&3 \end{array}} \right],B = \left[ {\begin{array}{c} 2&1 \\ 2&3 \end{array}} \right]$ and $C = \left[ {\begin{array}{c} { - 3}&1 \\ 2&0 \end{array}} \right]$ verify that $(AB)C = A(BC)$ and $(A + B)C = AC + BC$
Solution:

We have
$AB = \left[ {\begin{array}{c} 1&2 \\ { - 2}&3 \end{array}} \right] \times \left[ {\begin{array}{c} 2&1 \\ 2&3 \end{array}} \right] = \left[ {\begin{array}{c} {1.2 + 2.2}&{1.1 + 2.3} \\ { - 2.2 + 3.2}&{ - 2.1 + 3.3} \end{array}} \right] = \left[ {\begin{array}{c} 6&7 \\ 2&7 \end{array}} \right]$

$BC = \left[ {\begin{array}{c} 2&1 \\ 2&3 \end{array}} \right] \times \left[ {\begin{array}{c} { - 3}&1 \\ 2&0 \end{array}} \right] = \left[ {\begin{array}{c} { - 6 + 2}&{2 + 0} \\ { - 6 + 6}&{2 + 0} \end{array}} \right] = \left[ {\begin{array}{c} { - 4}&2 \\ 0&2 \end{array}} \right]$

$AC = \left[ {\begin{array}{c} 1&2 \\ { - 2}&3 \end{array}} \right] \times \left[ {\begin{array}{c} { - 3}&1 \\ 2&0 \end{array}} \right] = \left[ {\begin{array}{c} { - 3 + 4}&{1 + 0} \\ {6 + 6}&{ - 2 + 0} \end{array}} \right] = \left[ {\begin{array}{c} 1&1 \\ {12}&{ - 2} \end{array}} \right]$

$B + C = \left[ {\begin{array}{c} 2&1 \\ 2&3 \end{array}} \right] + \left[ {\begin{array}{c} { - 3}&1 \\ 2&0 \end{array}} \right] = \left[ {\begin{array}{c} {2 - 3}&{1 + 1} \\ {2 + 2}&{3 + 0} \end{array}} \right] = \left[ {\begin{array}{c} { - 1}&2 \\ 4&3 \end{array}} \right]$

(i)

$\begin{gathered} (AB)C = \left[ {\begin{array}{c} 6&7 \\ 2&7 \end{array}} \right] \times \left[ {\begin{array}{c} { - 3}&1 \\ 2&0 \end{array}} \right] = \left[ {\begin{array}{c} { - 18 + 14}&{6 + 0} \\ { - 6 + 14}&{2 + 0} \end{array}} \right] = \left[ {\begin{array}{c} { - 4}&6 \\ 8&2 \end{array}} \right] \hspace{1em} \\ \hspace{1em} \\ A(BC) = \left[ {\begin{array}{c} 1&2 \\ { - 2}&3 \end{array}} \right] \times \left[ {\begin{array}{c} { - 4}&2 \\ 0&2 \end{array}} \right] = \left[ {\begin{array}{c} { - 4 + 0}&{2 + 4} \\ {8 + 0}&{ - 4 + 6} \end{array}} \right] = \left[ {\begin{array}{c} { - 4}&6 \\ 8&2 \end{array}} \right] \hspace{1em} \\ \hspace{1em} \\ \end{gathered} $

Therefore,$(AB)C = A(BC)$

(ii)

$\begin{gathered} A(B + C) = \left[ {\begin{array}{c} 1&2 \\ { - 2}&3 \end{array}} \right] \times \left[ {\begin{array}{c} { - 1}&2 \\ 4&3 \end{array}} \right] = \left[ {\begin{array}{c} { - 1 + 8}&{2 + 6} \\ {2 + 12}&{ - 4 + 9} \end{array}} \right] = \left[ {\begin{array}{c} 7&8 \\ {14}&5 \end{array}} \right] \hspace{1em} \\ \hspace{1em} \\ AB + AC = \left[ {\begin{array}{c} 6&7 \\ 2&7 \end{array}} \right] + \left[ {\begin{array}{c} 1&1 \\ {12}&{ - 2} \end{array}} \right] = \left[ {\begin{array}{c} {6 + 1}&{7 + 1} \\ {2 + 12}&{7 - 2} \end{array}} \right] = \left[ {\begin{array}{c} 7&8 \\ {14}&5 \end{array}} \right] \hspace{1em} \\ \hspace{1em} \\ \end{gathered} $
Therefore,$(A + B)C = AC + BC$
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