Question:
If $A$ is a square matrix such that $A^{2}=A$, then $(I+A)^{3}-7 A$ is equal to
(a) $A$
(b) $I-A$
(c) $I$
(d) $3 \mathrm{~A}$
Solution:
(c) $I$
Here,
$A^{2}=A \quad \ldots(1)$
$A^{3}=A^{2} A$
$=A^{2} \quad$ [From eq. (1) $]$
$=A$
$\therefore A^{3}=A \quad \ldots(2)$
We know that $(I+A)^{3}=I^{3}+3(I)^{2} A+3(I) A^{2}+A^{3}$
$\Rightarrow(I+A)^{3}=I+3 A+3 A+A \quad$ [From eqs. (1) and (2)]
$\Rightarrow(I+A)^{3}=I+7 A$
$\Rightarrow(I+A)^{3}-7 A=I$
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