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

Question

Accepted Answer

"

$\mathrm{AA}^{\mathrm{T}}=\mathrm{I}=\mathrm{A}^{\mathrm{T}} \mathrm{A}$

On solving given expression, we get

$$\begin{aligned} & \frac{1}{2} \mathrm{~A}\left[\mathrm{~A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2+2 \mathrm{~A} \mathrm{~A}^{\mathrm{T}}+\mathrm{A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2-2 \mathrm{~A} \mathrm{~A}^{\mathrm{T}}\right] \\ & =\mathrm{A}\left[\mathrm{A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2\right]=\mathrm{A}^3+\mathrm{A}^{\mathrm{T}} \end{aligned}$$

"

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

Solution

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

Solution

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