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
$\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is D $2$

As the given system of equations has a non-trivial solution
$\Delta =\left|\begin{array}{ccc}p& a& a\\ b& q& b\\ c& c& r\end{array}\right|=0$

Applying $C_2\to C_2-C_1$ and $C_3\to C_3-C_2$

$\Delta =\left|\begin{array}{ccc}p& a-p& a-p\\ b& q-b& 0\\ c& 0& r-c\end{array}\right|=0$

Expanding along $C_3$, we get

$(a-p)\left|\begin{array}{cc}b& q-b\\ c& 0\end{array}\right|+(r-c)\left|\begin{array}{cc}p& a-p\\ b& q-b\end{array}\right|=0$

$\Rightarrow (a-p)(-c)(q-b)+(r-c)\left\{p\right(q-b)-b(a-p\left)\right\}=0$

$\Rightarrow (p-a)(q-b)c+p(r-c)(q-b)+b(r-c)(p-a)=0$

Dividing by $(p-a)(q-b)(r-c)$ we get

$\frac{c}{r-c}+\frac{p}{p-a}+\frac{b}{q-b}=0$

$\Rightarrow \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=\frac{q-b}{q-b}+\frac{r-c}{r-c}=2$