Question

If $p\ne a,q\ne b,r\ne c$ and the system of equations
$px+ay+az=0$
$bx+qy+bz=0$
$cx+cy+rz=0$
has a non-trivial solution, then the value of $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

Answer: (D)

As the given system of equations has a nontrivial solution
$\Delta =\left|\begin{array}{ccc}p& a& a\\ b& q& b\\ c& c& r\end{array}\right|=0$
Applying $C_3\to C_3-C_2$ and $C_2\to C_2-C_1$, we get
$\Delta =\left|\begin{array}{ccc}p& a-p& a-p\\ b& q-b& 0\\ c& 0& r-c<br/>\end{array}\right|=0$
Expanding along $C_3$, we get
$(a-p)\left|\begin{array}{cc}b& q-b\\ c& 0\end{array}\right|+(r-c)\left|\begin{array}{cc}p& a-p\\ b& q-b\end{array}\right|=0$
$\Rightarrow (a-p)(-c)(q-b)+(r-c)\{p(q-b)-b(a-p)\}=0$
$\Rightarrow (p-a)(q-b)c+p(r-c)(q-b)+b(r-c)(p-a)=0$
Dividing by $(p-a)(q-b)(r-c)$ we get
$\frac{c}{r-c}+\frac{p}{p-a}+\frac{b}{q-b}=0$
$\Rightarrow \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=\frac{q-b}{q-b}+\frac{r-c}{r-c}=2$