JEE MAIN - Mathematics (2024 - 29th January Morning Shift - No. 14)
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
$$\mathrm{A}^2+\mathrm{A}^{\mathrm{T}}$$
$$\mathrm{A}^3+\mathrm{I}$$
$$\mathrm{A}^3+\mathrm{A}^{\mathrm{T}}$$
$$\mathrm{A}^2+\mathrm{I}$$
Explanation
$$\mathrm{AA}^{\mathrm{T}}=\mathrm{I}=\mathrm{A}^{\mathrm{T}} \mathrm{A}$$
On solving given expression, we get
$$\begin{aligned} & \frac{1}{2} \mathrm{~A}\left[\mathrm{~A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2+2 \mathrm{~A} \mathrm{~A}^{\mathrm{T}}+\mathrm{A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2-2 \mathrm{~A} \mathrm{~A}^{\mathrm{T}}\right] \\ & =\mathrm{A}\left[\mathrm{A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2\right]=\mathrm{A}^3+\mathrm{A}^{\mathrm{T}} \end{aligned}$$
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