Answer:
Option B
Explanation:
We have , $a_{ij}=x_{j}+y_{i}$
$A= (a_{i})_{n xn}$
$A=\begin{bmatrix}x+y & 2x+y&3x+y&.....nx+y \\x+2y & 2x+2y&3x+2y &......nx+2y\\ x+3y&...&....&.....\\ x+ny &...&....&nx+ny \end{bmatrix}_{n\times n}$
$A=\begin{bmatrix}1 & 2&3......&n \\1 & 2&3..... &n\\ 1&...&....&.....\\ 1 &2&3.....&n \end{bmatrix}+y\begin{bmatrix}1 & 1&1.....&1 \\2 & ..&.. &2\\ 3&...&....&3\\ n &..&..&n \end{bmatrix}$
A=xP+yQ
where
$P=\begin{bmatrix}1 & 2&3&n \\1 & 2&3 &n\\ 1&2&....&.....\\ 1 &2&3&n \end{bmatrix}and Q\begin{bmatrix}1 & 1&1....&1 \\2 &2 &2.... &2\\ 3&...&....&3\\ ....n &..&n...&n \end{bmatrix}$
$\therefore$ (P+Q) is symmetric and (P-Q) is skew symmetric