MCQ
Let $A$ be a square matrix such that $A A^T=I$. Then $\frac{1}{2} A\left[\left(A+A^T\right)^2+\left(A-A^T\right)^2\right]$ is equal to
  • A
     $A^2+I$
  • B
    $A^3+I$
  • C
     $A^2+A^T$
  •  $A^3+A^T$

Answer

Correct option: D.
 $A^3+A^T$
d
$\mathrm{AA}^{\mathrm{T}}=\mathrm{I}=\mathrm{A}^{\mathrm{T}} \mathrm{A}$

On solving given expression, we get

$ \frac{1}{2} A\left[A^2+\left(A^T\right)^2+2 A A^T+A^2+\left(A^T\right)^2-2 A A^T\right] $

$ =A\left[A^2+\left(A^T\right)^2\right]=A^3+A^T$

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