First slide

Operations of matrix

Question

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

Easy

A

$A$
B

$I - A$
C

$I + A$
D

$3 A$

Solution

We have $A^{2} = I$
$∴ {(A-I)}^{3} {+(A+I)}^{3} -7A= [\{(A - I) + (A + I)\} \{{(A - I)}^{2} + {(A + I)}^{2} - (A - I) (A + I)\}] - 7 A$
$[∴ a^{3} {+b}^{3} {= (a+b)(a}^{2} {+b}^{2} -ab)]$
$\begin{array}{l} = [(2 A) \{2 (A^{2} + I^{2}) - (A^{2} - I^{2})\}] - 7 A \\ = 2 A [2 (I + I^{2}) - A^{2} + I^{2}] - 7 A [∵ A^{2} = I] \\ = 2 A [2 (I + I)] - 7 A (I^{2} = I) \\ = 8 A I - 7 A \\ = A \end{array}$

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