If A is a square matrix such that $A^{2} = I$ , then find the simplified value of $(A - I)^{3} + (A + I)^{3} - 7 A$ .

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Solution

$A^{2} = I$

$∴ A^{3} = A^{2} A = I A = A$

We know

$(A + B)^{3} = A^{3} + 3 A^{2} B + 3 A B^{2} + B^{3}$

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

Provided that $A B = B A$

Since, $A I = I A = A$

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

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

$\Rightarrow (A + I)^{3} = A^{3} + 3 A^{2} + 3 A + I$

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

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

$= 2 (A + 3 A)$

Hence, $= 8 A$

$(A - I)^{3} + (A + I)^{3} - 7 A = 8 A - 7 A = A$

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