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 ↔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.