Given A2=A and need to find (I+A)3−8I.
(I+A)2=(I+A)(I+A)
=I+A+A+A2
=I+2A+A2
=I+2A+A
=I+3A
(I+A)3=(I+A)2×(I+A)
=(I+3A)(I+A)
=I+A+3A+3A2
=I+4A+3A2
=I+4A+3A
=I+7A
(I+A)3−8I=(I+7A)−8I
=I+7A−8I
=7A−7I
=7(A−I)
Therefore, (I+A)3−8I=7(A−I)
If A is a square matrix and I is an identity matrix of same order such that A2=A, then (I+A)3−8I is equal to
Held on 22 May 2025 · Verified 13 Jul 2026.
A
A - I
7(A - I)
I
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