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

Options

A. I
B. 0
C. I-A
D. I+A

Answer

To solve the problem, we need to simplify the expression \((I - A)^3 + A\) given that \(A^2 = A\).

### Step-by-step Solution:

1. **Understand the Given Condition**:
We know that \(A^2 = A\). This means that \(A\) is an idempotent matrix.

2. **Expand the Expression**:
We need to expand \((I - A)^3\). We can use the binomial theorem or the identity for the cube of a binomial:
\[ (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \]
Here, let \(a = I\) and \(b = A\):
\[ (I - A)^3 = I^3 - 3I^2A + 3IA^2 - A^3 \]

3. **Substitute Known Values**:
Since \(I^3 = I\) and \(I^2 = I\), we can substitute these values:
\[ (I - A)^3 = I - 3IA + 3A^2 - A^3 \]
We also know \(A^2 = A\), so:
\[ (I - A)^3 = I - 3A + 3A - A^3 \]

4. **Simplify Further**:
We also know \(A^3 = A^2 \cdot A = A \cdot A = A\):
\[ (I - A)^3 = I - 3A + 3A - A \]
This simplifies to:
\[ (I - A)^3 = I - A \]

5. **Add A to the Expression**:
Now, we need to add \(A\) to \((I - A)^3\):
\[ (I - A)^3 + A = (I - A) + A \]
The \(A\) terms cancel out:
\[ (I - A)^3 + A = I \]

### Final Result:
Thus, we find that:
\[ (I - A)^3 + A = I \]