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# If $A$ and $B$ are square matrices of the same order such that $(A+B)(A-B)=A^2-B^2$ then $(ABA^{-1})^2$ is equal to

$(a)\;B^2\qquad(b)\;I\qquad(c)\;A^2B^2\qquad(d)\;A^2$

Can you answer this question?

Given :
$(A+B)(A-B)=A^2-B^2$
$A^2-AB+BA-B^2=A^2-B^2$
$AB=BA$
Now $(ABA^{-1})^2=(BAA^{-1})^2$
$\Rightarrow B^2$
Hence (a) is the correct answer.
answered Nov 22, 2013