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