To multiply two matrices, the number of columns in the first matrix must be equal to the number of rows in the second matrix. Consider the multiplications of 3×3 and 3×2 matrices.

**Multiplication of Matrices**

To multiply two matrices, the number of columns in the first
matrix must be equal to the number of rows in the second matrix. Consider the
multiplications of 3×3 and 3×2 matrices.

Matrices are multiplied by multiplying the elements in a row of
the first matrix by the elements in a column of the second matrix, and adding
the results

For example, product of matrices

The product *AB* can be found if the number of columns of matrix
*A* is equal to the number of rows of matrix *B*. If the order of
matrix *A* is *m* ×*n* and *B* is *n* × *p* then
the order of *AB* is *m* × *p*

If *A* is of order *m* ×*n* and *B* of the
order *n* × *p* then *AB* is defined but *BA* is not
defined. Even if *AB* and *BA* are both defined, it is not necessary
that they are equal. In general *AB* ≠ *BA*.

(i) If *A, B,
C* are *m *×*n *,* n *×* p *and
*n *×* p *matrices respectively then* A*(*B *+* C*)*
*=* AB *+* AC *(Right Distributive Property)

(ii) If *A,* *B, C* are *m
*×*n *,* m *×*n *and *n *×* p*
matrices respectively then(*A* + *B*)*C* = *AC* + *BC*
(Left Distributive Property)

If *A, B, C* are *m* × *n* , *n* × *p*
and *p* ×*q* matrices respectively then (*AB*)*C* = *A*(*BC*)

If *A* is a square matrix of order *n* ×*n* and *I*
is the unit matrix of same order then *AI *=* IA *=* A *.

**Note**

·
If** ***x*** **and** ***y*** **are two
real numbers such that** ***xy*** **=** **0** **then either**
***x*** **=** **0** **or** ***y*** **=** **0** **.
But this** **condition may not be true with respect to two matrices.

·
*AB*** **=** **0** **does not necessarily imply that** ***A***
**=** **0** **or** ***B*** **=** **0** **or both** ***A*,**
***B*** **=** **0

**Example 3.64**

If , find *AB*.* *

*Solution*

We observe that* **A*** **is a

**Example 3.65**

If find *AB*
and *BA*. Check if *AB = BA*.

*Solution*

We observe that* **A*** **is a

Therefore, AB ≠ BA.

Show that *A* and *B* satisfy commutative property with
respect to matrix multiplication.

We have to show that* **AB*** **=

Hence LHS = RHS (ie) *AB* = *BA*

**Example 3.67**

Substituting *y* = 2 in (1), 2*x* + 2 = 4 gives *x* = 1

Therefore, *x* = 1 , *y* = 2 .

**Note**

·
If *A *and *B *are any two non zero matrices, then (*A*
+ *B*)^{2} ≠ *A*^{2} + 2*AB* + *B*^{2.}

·
However if** ***AB*** **=** ***BA*** **then
(*A* + *B*)^{2} = *A*^{2} + 2*AB*
+ *B*^{2}

show that (*AB*)*C* = *A*(*BC*) .

*A*(*BC*)* *=* *(−1* *−* *4* *+*
*14 3 − 3 − 2) =(9−2) …(2)

From (1) and (2), (*AB*)*C* = *A*(*BC*) .

If verify that *A*(*B + C*) = *AB* + *AC*.

**Example 3.70**

If show that (AB)^{T} = B^{T}A^{T}

*Solution*

From (1) and (2),(*AB*)* ^{T}* =

Hence proved.

Tags : Example Solved Problem | Mathematics Example Solved Problem | Mathematics

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