If A is a square matrix such that

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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