If A and B are square matrices of the same order, explain, why in general
Question:

If A and B are square matrices of the same order, explain, why in general

(i) $(A+B)^{2} \neq A^{2}+2 A B+B^{2}$

(ii) $(A-B)^{2} \neq A^{2}-2 A B+B^{2}$

 

(iii) $(A+B)(A-B) \neq A^{2}-B^{2}$.

Solution:

(i) $\mathrm{LHS}=(A+B)^{2}$

$=(A+B)(A+B)$

$=A(A+B)+B(A+B)$

$=A^{2}+A B+B A+B^{2}$

We know that a matrix does not have commutative property. So,

AB ≠ BA
Thus,

$(A+B)^{2} \neq A^{2}+2 A B+B^{2}$

(ii) LHS $=(A-B)^{2}$

$=(A-B)(A-B)$

$=A(A-B)-B(A-B)$

 

$=A^{2}-A B-B A+B^{2}$

We know that a matrix does not have commutative property. So,
AB ≠ BA
Thus,

$(A-B)^{2} \neq A^{2}-2 A B+B^{2}$

$($ iii $) \mathrm{LHS}=(A+B)(A-B)$

$=A(A-B)+B(A-B)$

 

$=A^{2}-A B+B A-B^{2}$

We know that a matrix does not have commutative property. So,
AB ≠ BA
Thus,

$(A+B)(A-B) \neq A^{2}-B^{2}$

Administrator

Leave a comment

Please enter comment.
Please enter your name.