# If A is square ma

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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