Question

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

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

Answer 1

Answer: (C)

$2$

$A=\left|\begin{array}{ccc}p& q-y& r-z\\ p-x& q& r-z\\ p-x& q-y& r\end{array}\right|=0$
By operation of matrix (5),
$A=p(qr-(q-y\left)\right(r-z\left)\right)-(q-y)\left[\right(p-x)r-(p-x\left)\right(r-z\left)\right]+\left[\right(r-z\left)\right(p-x\left)\right(q-y)-(p-x\left)q\right]$
$=pqr-p(qr-qz-yr+yz)-(q-y)(pr-xr-pr+pz+xr-xy)+(r-z)(pq-py-xq+xy-pq+xq)$
$=pqz+pyr-pyz-(q-y)(pz-xy)+(r-z)(xy-py)$
$=pqz+pyr-pyz-qpz+qxz+pyz-xyz+rxy-rpy-xyz+pyz$
$A=qxz+rzy+pyz-2xyz$
So,$A=qxz+rxy+pyz-2xyz=0$

Dividing throughout by $xyz$,we get,
$\Rightarrow \frac{q}{y}+\frac{r}{z}+\frac{p}{x}=2$