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If A is a square matrix and I is an identity matrix of same order such that A2=AA^2 = A, then (I+A)3−8I(I + A)^3 - 8I is equal to

Solution

✅ Correct Option: 3

Given A2=AA^2 = A and need to find (I+A)3−8I(I + A)^3 - 8I.


(I+A)2=(I+A)(I+A)(I + A)^2 = (I + A)(I + A)

=I+A+A+A2= I + A + A + A^2

=I+2A+A2= I + 2A + A^2

=I+2A+A= I + 2A + A

=I+3A= I + 3A


(I+A)3=(I+A)2×(I+A)(I + A)^3 = (I + A)^2 \times (I + A)

=(I+3A)(I+A)= (I + 3A)(I + A)

=I+A+3A+3A2= I + A + 3A + 3A^2

=I+4A+3A2= I + 4A + 3A^2

=I+4A+3A= I + 4A + 3A

=I+7A= I + 7A


(I+A)3−8I=(I+7A)−8I(I + A)^3 - 8I = (I + 7A) - 8I

=I+7A−8I= I + 7A - 8I

=7A−7I= 7A - 7I

=7(A−I)= 7(A - I)


Therefore, (I+A)3−8I=7(A−I)(I + A)^3 - 8I = 7(A - I)

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