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`
D. `3A`

Explanation

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

### Step-by-step Solution:

1. **Expand \((I + A)^3\)**:
\[ (I + A)^3 = (I + A)(I + A)(I + A) \]
Using the binomial expansion:
\[ (I + A)^3 = I^3 + 3I^2A + 3IA^2 + A^3 \]
Since \(I^3 = I\) and \(A^2 = A\), we can simplify:
\[ = I + 3IA + 3A + A^3 \]

2. **Substituting \(A^3\)**:
We know that \(A^2 = A\), hence \(A^3 = A \cdot A^2 = A \cdot A = A\).
Therefore, we substitute \(A^3\) with \(A\):
\[ = I + 3IA + 3A + A \]
\[ = I + 3IA + 4A \]

3. **Substituting \(IA\)**:
Since \(IA = A\):
\[ = I + 3A + 4A \]
\[ = I + 7A \]

4. **Final Expression**:
Now we substitute this back into the original expression:
\[ (I + A)^3 - 7A = (I + 7A) - 7A \]
\[ = I \]

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