△=\(\begin{vmatrix} b+c & q+r & y+z\\ c+a & r+p & z+x\\ a+b&p+q&x+y \end{vmatrix}\)
=\(\begin{vmatrix} b+c & q+r & y+z\\ c+a & r+p & z+x\\ a&p&x \end{vmatrix}\)+\(\begin{vmatrix} b+c & q+r & y+z\\ c+a & r+p & z+x\\ b&q&y \end{vmatrix}\)
=△1+△2(say) ....(1)
Now △1= \(\begin{vmatrix} b+c & q+r & y+z\\ c+a & r+p & z+x\\ a&p&x \end{vmatrix}\)
Applying R2 → R2 − R3, we have:
△1=\(\begin{vmatrix} b+c & q+r & y+z\\ c & r & z\\ a&p&x \end{vmatrix}\)
Applying R1 → R1 − R2, we have:
△1=\(\begin{vmatrix} b & q & y\\ c & r & z\\ a&p&x \end{vmatrix}\)
Applying R1 ↔R3 and R2 ↔R3, we have:
△1=(-1)2\(\begin{vmatrix}a&p&x\\b&q&y\\c&r&z\end{vmatrix}=\begin{vmatrix}a&p&x\\b&q&y\\c&r&z\end{vmatrix}\) ….....(2)
△2=\(\begin{vmatrix} b+c & q+r & y+z\\ c+a & r+p & z+x\\ b&q&y \end{vmatrix}\)
Applying R1 → R1 − R3, we have:
△2=\(\begin{vmatrix} c & r & z\\ c+a & r+p & z+x\\ b&q&y \end{vmatrix}\)
Applying R2 → R2 − R1, we have:
△2=\(\begin{vmatrix} c & r & z\\ a & p & x\\ b&q&y \end{vmatrix}\)
Applying R1 ↔R2 and R2 ↔R3, we have:
△2=(-1)2\(\begin{vmatrix} a & p& x\\ b & q & y\\ c&r&z \end{vmatrix}\)= \(\begin{vmatrix} a & p& x\\ b & q & y\\ c&r&z \end{vmatrix}\) ...(3)
From (1), (2), and (3), we have:
△=2\(\begin{vmatrix} a & p& x\\ b & q & y\\ c&r&z \end{vmatrix}\) Hence, the given result is proved.
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Read More: Properties of Determinants