3 and 4 .Determinants and Matrices
medium

If $A$ is square matrix such that $A^{2}=A$, then $(1+A)^{3}-7 A$ is equal to

A

$A$

B

$I-A$

C

$3A$

D

$I$

Solution

$(I+A)^{3}-7 A=I^{3}+A^{3}+3 I^{2} A+3 A^{2} I-7 A$

$=I+A^{3}+3 A+3 A^{2}-7 A$

$=I+A^{2} \cdot A+3 A+3 A-7 A$                       $\left[A^{2}=A\right]$

$=I+A \cdot A-A$

$=I+A^{2}-A$

$=I+A-A$

$=I$

$\therefore(I+A)^{3}-7 A=I$

Standard 12
Mathematics

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