Question

Using the property of determinants and without expanding,prove that:\(\begin{vmatrix} b+c & q+r & y+z\\ c+a & r+p & z+x\\ a+b&p+q&x+y \end{vmatrix}\)=2\(\begin{vmatrix} a & p & x\\ b & q & y\\c&r&z \end{vmatrix}\)

Answer 1

△=\(\begin{vmatrix} b+c & q+r & y+z\\ c+a & r+p & z+x\\ a+b&p+q&x+y \end{vmatrix}\)

=\(\begin{vmatrix} b+c & q+r & y+z\\ c+a & r+p & z+x\\ a&p&x \end{vmatrix}\)+\(\begin{vmatrix} b+c & q+r & y+z\\ c+a & r+p & z+x\\ b&q&y \end{vmatrix}\) 
=△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 ↔R_3, 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 − R_1, 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.