Question:
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) $3 \mathrm{~A}$
Solution:
$(A-I)^{3}+(A+I)^{3}-7 A$
$=A^{3}-I^{3}-3 A^{2} I+3 A I^{2}+A^{3}+I^{3}+3 A^{2} I+3 A I^{2}-7 A$
$=2 A^{3}+6 A I^{2}-7 A$
$=2 A \cdot A^{2}+6 A-7 A$
$=2 A \cdot I-A \quad\left(\because A^{2}=I\right)$
$=2 A-A$
$=A$
Hence, the correct option is (a).
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