If `"AA"^T=I` where `A^T` is transpose of matrix A then, `1/2A[(A+A^T)^2+(A-A^T)^2]=?`

Options

A. `A^2`
B. `A^3+1`
C. `A^2+1`
D. `A^3+A^T`

Answer

To solve the problem \( \frac{1}{2} A \left[ (A + A^T)^2 + (A - A^T)^2 \right] \) given that \( AA^T = I \), we will follow these steps:

### Step 1: Expand the expressions
We start by expanding \( (A + A^T)^2 \) and \( (A - A^T)^2 \).

1. **Expanding \( (A + A^T)^2 \)**:
\[ (A + A^T)^2 = A^2 + AA^T + A^TA + (A^T)^2 \]

2. **Expanding \( (A - A^T)^2 \)**:
\[ (A - A^T)^2 = A^2 - AA^T - A^TA + (A^T)^2 \]

### Step 2: Combine the expanded expressions
Now, we can combine the two expansions:
\[ (A + A^T)^2 + (A - A^T)^2 = \left( A^2 + AA^T + A^TA + (A^T)^2 \right) + \left( A^2 - AA^T - A^TA + (A^T)^2 \right) \]

### Step 3: Simplify the combined expression
When we combine the two expansions, we notice that the \( AA^T \) and \( A^TA \) terms cancel each other out:
\[ = 2A^2 + 2(A^T)^2 \]

### Step 4: Substitute back into the original expression
Now, we substitute this back into the original expression:
\[ \frac{1}{2} A \left[ 2A^2 + 2(A^T)^2 \right] = A \left[ A^2 + (A^T)^2 \right] \]

### Step 5: Use the given condition \( AA^T = I \)
Since \( AA^T = I \), we can replace \( A^TA \) with \( I \):
\[ = A \left[ A^2 + I \right] \]

### Step 6: Final expression
Thus, the final expression simplifies to:
\[ = A^3 + A \]

### Conclusion
The final answer is:
\[ A^3 + A \]