Question

If $p \neq a, q \neq b, r \neq c$ and the system of equations
$p x+a y+a z =0 $
$b x+q y+b z =0 $
$c x+c y+r z =0$
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
$\Delta=\begin{vmatrix}p & a & a \\b & q & b \\c & c & r\end{vmatrix}=0$
Applying $C_3 \rightarrow C_3-C_2$ and $C_2 \rightarrow C_2-C_1$, we get
$\Delta=\begin{vmatrix}p & a-p & a-p \\b & q-b & 0 \\c & 0 & r-c
\end{vmatrix}=0$
Expanding along $C_3$, we get
$(a-p)\begin{vmatrix}b & q-b \\c & 0\end{vmatrix}+(r-c)\begin{vmatrix}p & a-p \\b & q-b\end{vmatrix}=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 $
$\Rightarrow \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=\frac{q-b}{q-b}+\frac{r-c}{r-c}=2$