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

A

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I + A

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3 A

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I - A

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Solution

The correct option is A $A$
We can expand $(A + I)^{n}$ using the expansion of $(a + b)^{n}$ , where $n \in N$
$∴ (A - I)^{3} + (A + I)^{3} - 7 A$
$= A^{3} - 3 A^{2} I + 3 A I^{2} - I^{3} + A^{3} + 3 A^{2} I + 3 A I^{2} + I^{3} - 7 A$
$= A^{3} - 3 A^{2} + 3 A - I^{3} + A^{3} + 3 A^{2} + 3 A + I^{3} - 7 A$
$[∵ A \cdot I = A = I and I^{2} = I]$
$= 2 A^{3} + 6 A - 7 A$
$= 2 A^{2} A + 6 A - 7 A$
$= 2 I A + 6 A - 7 A [Given A^{2} = I]$
$= 2 A + 6 A - 7 A$
$= A$
Hence option (a) is correct.

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