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+AT
• D. A3+AT

Answer 1

Answer: (D)

A3+AT

Answer 2

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: $A^3 + A^\top$