Question

The value of the determinant $\begin{vmatrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{vmatrix} \text { is }$

• A. $2\begin{vmatrix}a & p & x \\ b & q & y \\ c & r & z\end{vmatrix}$
• B. $\begin{vmatrix}a & p & x \\ b & q & y \\ c & r & z\end{vmatrix}$
• C. $-\begin{vmatrix}a & p & x \\ b & q & y \\ c & r & z\end{vmatrix}$
• D. 0

Answer 1

Answer: (A)

Here, $\begin{vmatrix}b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y\end{vmatrix}$
$=\begin{vmatrix}b+c & c+a & a+b \\ q+r & r+p & p+q \\ y+z & z+x & x+y\end{vmatrix}$
(interchange row and column)
$=\begin{vmatrix}b+c & c+a & -2 c \\ q+r & r+p & -2 r \\ y+z & z+x & -2 z\end{vmatrix}$
using $C_3 \rightarrow C_3-\left(C_1+C_2\right)$
$=-2\begin{vmatrix}b+c & c+a & c \\ q+r & r+p & r \\ y+z & z+x & z\end{vmatrix}$
(taking -2 common from $C_3$ )
$=-2\begin{vmatrix}b & a & c \\ q & p & r \\ y & x & z\end{vmatrix}$
(using $C_1 \rightarrow C_1-C_3$ and $C_2 \rightarrow C_2-C_3$ )
$=2\begin{vmatrix}a & b & c \\ p & q & r \\ x & y & z\end{vmatrix} $ (using $C_1 \leftrightarrow C_2$ )
$=2\begin{vmatrix}a & p & x \\b & q & y \\c & r & z\end{vmatrix}$
(interchange column and row)