If `A^2 = A`, then `(I + A)^(4)` is equal to

Options

A. `I + 15A`
B. `1 + 7A`
C. `1 + 8A`
D. `1 + 11A`

Answer

To solve the problem, we need to find \((I + A)^4\) given that \(A^2 = A\).

### Step-by-Step Solution:

1. **Understanding the Property of Matrix A**:
Since \(A^2 = A\), we can conclude that \(A\) is an idempotent matrix. This means that multiplying \(A\) by itself gives back \(A\).

2. **Expressing \((I + A)^4\)**:
We can express \((I + A)^4\) as:
\[ (I + A)^4 = (I + A)^2 \cdot (I + A)^2 \]
So, we first need to find \((I + A)^2\).

3. **Calculating \((I + A)^2\)**:
Using the formula for the square of a binomial:
\[ (I + A)^2 = I^2 + 2IA + A^2 \]
Since \(I^2 = I\) and \(A^2 = A\), we substitute these values:
\[ (I + A)^2 = I + 2A + A = I + 3A \]

4. **Finding \((I + A)^4\)**:
Now, we substitute \((I + A)^2\) back into our expression for \((I + A)^4\):
\[ (I + A)^4 = (I + 3A)^2 \]
Again, using the binomial expansion:
\[ (I + 3A)^2 = I^2 + 2(3A)I + (3A)^2 = I + 6A + 9A^2 \]
Now substituting \(A^2 = A\):
\[ (I + 3A)^2 = I + 6A + 9A = I + 15A \]

5. **Final Result**:
Thus, we conclude that:
\[ (I + A)^4 = I + 15A \]

### Final Answer:
\[ (I + A)^4 = I + 15A \]