If A is a square matrix such that


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

(a) $A$

(b) $I-A$

(c) $I+A$

(d) $3 \mathrm{~A}$


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

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

$=2 A^{3}+6 A I^{2}-7 A$

$=2 A \cdot A^{2}+6 A-7 A$

$=2 A \cdot I-A \quad\left(\because A^{2}=I\right)$

$=2 A-A$


Hence, the correct option is (a).

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