Question

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

• A. A²+I
• B. A³+I
• C. A²+A^T
• D. A³+A^T

Answer 1

The Correct Option is D

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: \(A^3 + A^\top\)