2 Express the matrix $\left[\begin{array}{ccc}2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2\end{array}\right]$ as the sum of a symmetric and a skew-symmetric matrix.
$$\begin{aligned} &\text { We have, }\\ &\begin{aligned} & A=\left[\begin{array}{ccc} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{array}\right] \\ \therefore\quad & A^{\prime}=\left[\begin{array}{ccc} 2 & 1 & 4 \\ 3 & -1 & 1 \\ 1 & 2 & 2 \end{array}\right] \end{aligned} \end{aligned}$$
$$\begin{aligned} \text{Now,}\quad & \frac{A+A^{\prime}}{2}=\frac{1}{2}\left[\begin{array}{ccc} 4 & 4 & 5 \\ 4 & -2 & 3 \\ 5 & 3 & 4 \end{array}\right]=\left[\begin{array}{ccc} 2 & 2 & \frac{5}{2} \\ 2 & -1 & \frac{3}{2} \\ \frac{5}{2} & \frac{3}{2} & 2 \end{array}\right] \\ \text{and}\quad & \frac{A-A^{\prime}}{2}=\frac{1}{2}\left[\begin{array}{ccc} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & \frac{-3}{2} \\ -1 & 0 & \frac{1}{2} \\ \frac{3}{2} & \frac{-1}{2} & 0 \end{array}\right] \end{aligned}$$
$$\begin{aligned} &\therefore \quad \frac{A+A^{\prime}}{2}+\frac{A-A^{\prime}}{2}=\left[\begin{array}{ccc} 2 & 2 & \frac{5}{2} \\ 2 & -1 & \frac{3}{2} \\ \frac{5}{2} & \frac{3}{2} & 2 \end{array}\right]+\left[\begin{array}{ccc} 0 & 1 & \frac{-3}{2} \\ -1 & 0 & \frac{1}{2} \\ \frac{3}{2} & \frac{-1}{2} & 0 \end{array}\right]\\ &\text { which is the required expression. } \end{aligned}$$
The matrix $P=\left[\begin{array}{lll}0 & 0 & 4 \\ 0 & 4 & 0 \\ 4 & 0 & 0\end{array}\right]$ is a
Total number of possible matrices of order $3\times3$ with each entry 2 or 0 is
$\left[\begin{array}{ll}2 x+y & 4 x \\ 5 x-7 & 4 x\end{array}\right]=\left[\begin{array}{cc}7 & 7 y-13 \\ y & x+6\end{array}\right]$, then the value of $x+y$ is
If $A=\frac{1}{\pi}\left[\begin{array}{cc}\sin ^{-1}(x \pi) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi x)\end{array}\right]$ and $B=\frac{1}{\pi}\left[\begin{array}{cc}-\cos ^{-1}(x \pi) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & -\tan ^{-1}(\pi x)\end{array}\right]$, then $A-B$ is equal to