Q47 of 146 Page 52

If A is square matrix such that A2 = A, show that (I + A)3 = 7A + I.

Given that,


A2 = A


∵ (a+b)3 = a3 + b3 + 3a2b + 3ab2


As, (I + A)3 = I3 + A3 + 3I2A + 3IA2


∵ I is an identity matrix.


∴ I3 = I2 = I


∴ (I + A)3 = I + A3 + 3IA + 3IA


As, I is an identity matrix.


∴ IA = AI = A


⇒ (I + A)3 = I + A3 + 6IA


∵ A2 = A


⇒ (I + A)3 = I + A2.A + 6A


⇒ (I + A)3 = I + A.A + 6A


⇒ (I + A)3 = I + A2 + 6A


⇒ (I + A)3 = I + A + 6A = I + 7A


Hence,


(I + A)3 = I + 7A …proved


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