If `A` is a square matrix such that `A^2=A ,` then `(I+A)^3-7A` is equal to

Options

A. A
B. I-A
C. I
D. 3A

Explanation

To solve the problem, we need to evaluate the expression \((I + A)^3 - 7A\) given that \(A\) is a square matrix such that \(A^2 = A\).

### Step-by-Step Solution:

1. **Expand \((I + A)^3\)**:
We can use the binomial theorem or the formula for the cube of a sum:
\[ (A + B)^3 = A^3 + B^3 + 3A^2B + 3AB^2 \]
Here, let \(A = I\) and \(B = A\):
\[ (I + A)^3 = I^3 + A^3 + 3I^2A + 3IA^2 \]

2. **Calculate each term**:
- \(I^3 = I\)
- Since \(A^2 = A\), we have \(A^3 = A \cdot A^2 = A \cdot A = A\).
- \(3I^2A = 3IA = 3A\)
- \(3IA^2 = 3A\) (since \(A^2 = A\))

Putting it all together:
\[ (I + A)^3 = I + A + 3A + 3A = I + 7A \]

3. **Subtract \(7A\)**:
Now we substitute this back into our expression:
\[ (I + A)^3 - 7A = (I + 7A) - 7A \]

4. **Simplify**:
The \(7A\) terms cancel out:
\[ (I + 7A) - 7A = I \]

Thus, the final result is:
\[ (I + A)^3 - 7A = I \]

### Final Answer:
The expression \((I + A)^3 - 7A\) is equal to \(I\).