Question

$If\left|\begin{array}{ccc}p& q-y& r-z\\ p-x& q& r-z\\ p-x& q-y& r\end{array}\right|=0$ and $x,y,z$ are non-zero then the value of $\frac{p}{x}+\frac{q}{y}+\frac{r}{z}$ is:

• A. $0$
• B. $1$
• C. $2$
• D. $4pqr$

Answer 1

The correct option is C $2$

$\left|\begin{array}{ccc}p& q-y& r-z\\ p-x& q& r-z\\ p-x& q-y& r\end{array}\right|=0$

Using vertex operations,

$R_2\to R_2-R_1$ & $R_3\to R_3-R_1$

$\left|\begin{array}{ccc}p& q-y& r-z\\ p-x-q& q-q+y& r-z-r+z\\ p-x-p& q-y-q+y& r-r+z\end{array}\right|=0$

$\left|\begin{array}{ccc}p& q-y& r-z\\ -x& y& 0\\ -x& 0& z\end{array}\right|=0$

on expanding this $-x(-(r-z\left)\right(y\left)\right)+z\left(py\right(-x\left)\right(q-y)$

$0\Rightarrow x\left(\right(r-z\left)y\right)+z(py+x(q-y\left)\right)$

$0\Rightarrow xyr-xyz+zyp+xzq-xyz=0$

Dividing $RHS$ by $\left(xyz\right)$

$\therefore 0=\frac{xyz}{xyz}-\frac{xyz}{xyz}+\frac{zyp}{xyz}+\frac{xzq}{xyz}-\frac{xyz}{xyz}$

$0=\frac{r}{z}-1+\frac{p}{x}+\frac{v}{y}-1$

or $\frac{p}{x}+\frac{q}{y}+\frac{r}{2}=2$

Option $C$