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

5 I

The correct answer is Option (1) → 5 I

Given: $A^2 = A$ and $I$ is identity matrix of same order

Let $M = (2I + A)^3 - 19A - 3I$

Since $A^2 = A$, then $A^3 = A^2 \cdot A = A \cdot A = A$

Now expand $(2I + A)^3$ using binomial expansion:

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

Using $A^2 = A$ and $A^3 = A$:

$(2I + A)^3 = 8I + 12A + 6A + A = 8I + 19A$

Now compute:

$M = (8I + 19A) - 19A - 3I = (8I - 3I) + (19A - 19A) = 5I$