Let A be a square matrix such that \(AA^T=I\).Then \(\frac{1}{2}A[(A+A^T)^2+(A-A^T)^2]\) is equal to
- A2+I
- A3+I
- A2+AT
- A3+AT
The Correct Option is D
Solution and Explanation
Step 1: Use the Condition \(AA^\top = I = A^\top A\)
Given that \(AA^\top = I\), we can substitute this property in the expression.
Step 2: Expand the Expression
\[ \frac{1}{2} A \left[(A + A^\top)^2 + (A - A^\top)^2\right] \]
Expanding \((A + A^\top)^2\) and \((A - A^\top)^2\), we get:
\[ \frac{1}{2} A \left[A^2 + (A^\top)^2 + 2AA^\top + A^2 + (A^\top)^2 - 2AA^\top\right] \]
\[ = A \left[A^2 + (A^\top)^2\right] \]
Step 3: Simplify the Result
\[ = A^3 + A^\top \]
So, the correct answer is: \(A^3 + A^\top\)
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