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

Options

A. A
B. I+A
C. I-A
D. I

Answer

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

### Step-by-Step Solution:

1. **Expand \((I + A)^3\)**:
We can use the binomial expansion formula for \((x + y)^3\):
\[ (I + A)^3 = I^3 + 3I^2A + 3IA^2 + A^3 \]

2. **Substituting values**:
Since \(I^3 = I\) and \(I^2 = I\), we can substitute these into the equation:
\[ (I + A)^3 = I + 3IA + 3IA^2 + A^3 \]

3. **Using the property \(A^2 = A\)**:
We know \(A^2 = A\), so we can replace \(IA^2\) with \(A\) and \(A^3\) with \(A^2 \cdot A = A\):
\[ (I + A)^3 = I + 3A + 3A + A = I + 5A \]

4. **Now substitute back into the original expression**:
We need to compute \((I + A)^3 - 7A\):
\[ (I + A)^3 - 7A = (I + 5A) - 7A \]

5. **Simplifying the expression**:
Combine like terms:
\[ (I + 5A - 7A) = I - 2A \]

### Final Result:
Thus, the expression \((I + A)^3 - 7A\) simplifies to:
\[ \boxed{I - 2A} \]