Question:
If $A$ is square matrix such that $A^{2}=A$, then $(I+A)^{3}-7 A$ is equal to
A. A
B. I − A
C. I
D. 3A
Solution:
Answer: C
$(I+A)^{3}-7 A=I^{3}+A^{3}+3 I^{2} A+3 A^{2} I-7 A$
$=I+A^{3}+3 A+3 A^{2}-7 A$
$=I+A^{2} \cdot A+3 A+3 A-7 A \quad\left[A^{2}=A\right]$
$=I+A \cdot A-A$
$=I+A^{2}-A$
$=I+A-A$
$=I$
$\therefore(I+A)^{3}-7 A=I$
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