Question

If $a \neq p, b \neq q, c \neq r$ and $\begin{vmatrix}p & b & c \\ a & q & c \\ a & b & r\end{vmatrix}=0$ then the value of $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is equal to

• A. -1
• B. 1
• C. -2
• D. 2

Answer 1

Answer: (D)

Given
$\begin{vmatrix}p & b & c \\ a & q & c \\ a & b & r\end{vmatrix}=0$
$R_{1} \rightarrow R_{1}-R_{2}, R_{2} \rightarrow R_{2}-R_{3}$ reduces the determinant to $\begin{vmatrix}p-a & b-q & 0 \\ 0 & q-b & c-r \\ a & b & r\end{vmatrix}=0$ $\Rightarrow(p-a)(q-b) r+a(b-q)(c-r)-b(p-a)(c-r)=0$
$\Rightarrow$ Dividing throughout by $(p-a)(q-b)(r-c)$, we get
$\frac{r}{r-c}+\frac{a}{p-a}+\frac{b}{q-b}=0$
$\Rightarrow \frac{r}{r-c}+1+\frac{a}{p-a}+1+\frac{b}{q-b}=2$
$\Rightarrow \frac{r}{r-c}+\frac{p}{p-a}+\frac{q}{q-b}=2$