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(B) Given $a_{i j} = xi + yj$. The matrix $A$ is defined as $A = [a_{i j}]_{n 	imes n}$.We can write $A$ as:$A = \begin{bmatrix} x+y & 2x+y & \dots &

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$P+Q$ is symmetric and $P-Q$ is skew-symmetric

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(B) Given $a_{i j} = xi + yj$. The matrix $A$ is defined as $A = [a_{i j}]_{n 	imes n}$.We can write $A$ as:$A = \begin{bmatrix} x+y & 2x+y & \dots & nx+y \ x+2y & 2x+2y & \dots & nx+2y \ \vdots & \vdots & \ddots & \vdots \ x+ny & 2x+ny & \dots & nx+ny \end{bmatrix}$This can be decomposed into:$A = x \begin{bmatrix} 1 & 2 & \dots & n \ 1 & 2 & \dots & n \ \vdots & \vdots & \ddots & \vdots \ 1 & 2 & \dots & n \end{bmatrix} + y \begin{bmatrix} 1 & 1 & \dots & 1 \ 2 & 2 & \dots & 2 \ \vdots & \vdots & \ddots & \vdots \ n & n & \dots & n \end{bmatrix}$Let $P = \begin{bmatrix} 1 & 2 & \dots & n \ 1 & 2 & \dots & n \ \vdots & \vdots & \ddots & \vdots \ 1 & 2 & \dots & n \end{bmatrix}$ and $Q = \begin{bmatrix} 1 & 1 & \dots & 1 \ 2 & 2 & \dots & 2 \ \vdots & \vdots & \ddots & \vdots \ n & n & \dots & n \end{bmatrix}$.Note that $P$ has identical rows, so its rank is $1$. $Q$ has proportional rows, so its rank is $1$.However, the question asks for properties of $P+Q$ and $P-Q$. $P+Q = [i+j]_{n 	imes n}$, which is a symmetric matrix because $(P+Q)^T = [j+i]^T = [i+j] = P+Q$.$P-Q = [i-j]_{n 	imes n}$, which is a skew-symmetric matrix because $(P-Q)^T = [j-i]^T = [-(i-j)] = -(P-Q)$.

If $x, y$ are any two non-zero real numbers, $a_{i j} = xi + yj$, $A = \{a_{i j}\}_{n \times n}$ and $P, Q$ are two $n \times n$ matrices such that $A = xP + yQ$, then

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