The value of $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ for given determinant is

Question

If $a\neq p$, $b\neq q$, $c\neq r$ and $\left|\begin{array}{cc}p&b&c\\a&q&c\\a&b&r \end{array}\right|= 0$ then the value of $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is

(a) -1 $\space\space\space\space\space$ (b)1 $\space\space\space\space\space$ (c)-2 $\space\space\space\space\space$ (d)2

I tried this question, but I could not go beyond

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

after applying two elementary row operations $(R_1 \rightarrow R_1 - R_2$ & $R_2 \rightarrow R_2 - R_3)$to the determinant.

Any idea on how to solve it further? Thanks...

Answer 1

Once you got $$(p-a)(r(q-b)+b(r-c))+a(q-b)(r-c)=0$$ Then divide this whole thing by $(p-a)(q-b)(r-c)$ to get $$\frac{r}{r-c}+\frac{b}{q-b}+\frac{a}{p-a}=0.$$ Now \begin{align*} \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c} & =\frac{p}{p-a}+\frac{q}{q-b}-\frac{b}{q-b}-\frac{a}{p-a}\\ & =\left[\frac{p}{p-a}-\frac{a}{p-a}\right]+\left[\frac{q}{q-b}-\frac{b}{q-b}\right]\\ &=2. \end{align*}