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$A^2+A^T$
- ✓$A^3+A^T$
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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$f(x)=\left\{\begin{array}{cc}x-[x] & \text { if }[x] \text { is odd } \\ 1+[x]-x & \text { if }[x] \text { is even }\end{array}\right.$
Then the value of $\frac{\pi^2}{10} \int_{-10}^{10} f(x) \cos \pi x d x$ is