Question:
If $A$ is a square matrix such that $A^{2}=I$, then $A^{-1}$ is equal to
(a) $A+I$
(b) $A$
(c) 0
(d) $2 \mathrm{~A}$
Solution:
(b) $A$
Given : $A^{2}=I$
On multiplying both sides by $A^{-1}$, we get
$A^{-1} A^{2}=A^{-1} I$
$\Rightarrow A=A^{-1} I$
$\Rightarrow A=A^{-1}$
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