Question:
If $A$ and $B$ are square matrices of the same order such that $A B=B A$, then show that $(A+B)^{2}=A^{2}+2 A B+B^{2}$.
Solution:
$(A+B)^{2}=(A+B)(A+B)$
$=A^{2}+A B+B A+B^{2}$
$=A^{2}+2 A B+B^{2}$ $(\because A B=B A)$
Hence, $(A+B)^{2}=A^{2}+2 A B+B^{2}$.
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