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