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# If A is a square matrix such that $A^2=I$,then $(A-I)^3+(A+I)^3-7A$ is equal to $(A)\quad A\quad (B)\quad I-A\quad (C)\quad I +A\quad(D)\quad 3A$

Given:$A^2=I$

$(A-I)^3+(A+I)^3-7A.$

We know that

$(a+b)^3=a^3+b^3+3ab(a+b)$

$(a-b)^3=a^3+b^3-3ab(a+b)$

$(A-I)^3+(A+I)^3-7A.$

$A^3-I^3-3AI(A-I)+A^3+I^3+3AI(A+I)-7A.$

$I.A-I^3-3A^2I+3AI^2+IA+I^3+3A^2I+3AI^2-7A.$

$A-I^3-3I^3+3AI^2+IA+I^3+3I^3+3AI^2-7A$

$A+3A+A+3A-7A$

$8A-7A=A.$

Hence the answer is option (A).