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If A is a square matrix such that A2=AA^2 = A and I is the identity matrix of same order as A, then the matrix (2I+A)3−19A−3I(2I+A)^3 - 19A - 3I is equal to

Solution

✅ Correct Option: 1

Given A2=AA^2 = A and II is the identity matrix of the same order as AA.

Since A2=AA^2 = A:

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

So every power of AA equals AA itself.


Expanding (2I+A)3(2I + A)^3 using the binomial formula:

(2I+A)3=(2I)3+3(2I)2(A)+3(2I)(A2)+A3(2I + A)^3 = (2I)^3 + 3(2I)^2(A) + 3(2I)(A^2) + A^3

(2I)3=8I(2I)^3 = 8I

3(2I)2(A)=3⋅4I⋅A=12A3(2I)^2(A) = 3 \cdot 4I \cdot A = 12A

3(2I)(A2)=6I⋅A2=6A3(2I)(A^2) = 6I \cdot A^2 = 6A      (since A2=AA^2 = A)

A3=AA^3 = A      (since A3=AA^3 = A)

So:

(2I+A)3=8I+12A+6A+A(2I + A)^3 = 8I + 12A + 6A + A

=8I+19A= 8I + 19A


(2I+A)3−19A−3I(2I+A)^3 - 19A - 3I

=(8I+19A)−19A−3I= (8I + 19A) - 19A - 3I

=8I+19A−19A−3I= 8I + 19A - 19A - 3I

=5I= 5I

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