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# If $A=\begin{vmatrix}a&b&c\\x&y&z\\p&q&r\end{vmatrix}$ and $B=\begin{vmatrix}q&-b&y\\-p&a&-x\\r&-c&z\end{vmatrix}$ then

$\begin{array}{1 1}(A)\;|A|=|B|&(B)\;|A|=-|B|\\(C)\;|A|=2|B|&(D)\;\text{None of these}\end{array}$

$|B|=-\begin{vmatrix}q&-b&y\\p&-a&x\\r&-c&z\end{vmatrix}$
$\Rightarrow \begin{vmatrix}q&b&y\\p&a&x\\r&c&z\end{vmatrix}$
$\Rightarrow -\begin{vmatrix} p&a&x\\q&b&y\\r&c&z\end{vmatrix}$
$\Rightarrow \begin{vmatrix}a&p&x\\b&q&y\\c&r&z\end{vmatrix}=\begin{vmatrix} a&b&c\\p&q&r\\x&y&z\end{vmatrix}$
$\Rightarrow -\begin{vmatrix}a&b&c\\x&y&z\\p&q&r\end{vmatrix}$
$\Rightarrow |B|=-|A|$
Hence (B) is the correct answer.