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If A is a square matrix and I is the identity matrix of same order such that A2=IA^2 = I, then 3(A−I)3+3(A+I)3−15A3(A - I)^3 + 3(A + I)^3 - 15A is equal to

Solution

✅ Correct Option: 2

There is a useful identity for adding two cubes with opposite signs:

(x+y)3+(x−y)3=2x3+6xy2(x + y)^3 + (x - y)^3 = 2x^3 + 6xy^2

Since II commutes with every matrix, this identity applies here.


Applying with x=Ax = A, y=Iy = I:

(A+I)3+(A−I)3=2A3+6A⋅I2(A + I)^3 + (A - I)^3 = 2A^3 + 6A \cdot I^2

Since I2=II^2 = I:

(A+I)3+(A−I)3=2A3+6A(A + I)^3 + (A - I)^3 = 2A^3 + 6A


Substituting into the original expression:

3[(A−I)3+(A+I)3]−15A3\left[(A - I)^3 + (A + I)^3\right] - 15A

=3[2A3+6A]−15A= 3\left[2A^3 + 6A\right] - 15A

=6A3+18A−15A= 6A^3 + 18A - 15A

=6A3+3A= 6A^3 + 3A


Since A2=IA^2 = I:

A3=A⋅A2=A⋅I=AA^3 = A \cdot A^2 = A \cdot I = A

So,

6A3+3A=6A+3A=9A6A^3 + 3A = 6A + 3A = \boxed{9A}

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