Matrices Question 6
Question 6 - 2024 (29 Jan Shift 1)
Let $A$ be a square matrix such that $A A^{T}=I$. Then $\frac{1}{2} \mathrm{~A}\left[\left(\mathrm{~A}+\mathrm{A}^{\mathrm{T}}\right)^{2}+\left(\mathrm{A}-\mathrm{A}^{\mathrm{T}}\right)^{2}\right]$ is equal to
(1) $\mathrm{A}^{2}+\mathrm{I}$
(2) $A^{3}+I$
(3) $A^{2}+A^{T}$
(4) $A^{3}+A^{T}$
Show Answer
Answer (4)
Solution
$\mathrm{AA}^{\mathrm{T}}=\mathrm{I}=\mathrm{A}^{\mathrm{T}} \mathrm{A}$
On solving given expression, we get
$$ \begin{aligned} & \frac{1}{2} \mathrm{~A}\left[\mathrm{~A}^{2}+\left(\mathrm{A}^{\mathrm{T}}\right)^{2}+2 \mathrm{AA}^{\mathrm{T}}+\mathrm{A}^{2}+\left(\mathrm{A}^{\mathrm{T}}\right)^{2}-2 \mathrm{AAA}^{\mathrm{T}}\right] \ & =\mathrm{A}\left[\mathrm{A}^{2}+\left(\mathrm{A}^{\mathrm{T}}\right)^{2}\right]=\mathrm{A}^{3}+\mathrm{A}^{\mathrm{T}} \end{aligned} $$
