Question

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

Answer 1

Applying $R_1-R_2$ and $R_2-R_3$, we have kept $p-a,q-b$ and $r-c$ as desired

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

Expand w.r.t. $C_1$.

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

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

Dividing throughout by $(p-a)(q-b)(r-c)$ which is justified

since $a\ne p,b\ne q,c\ne r$ is given

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

Add $1+1$ on both sides to get the desired form

$\frac{r}{r-c}+(\frac{b}{q-b}+1)+(\frac{a}{p-a}+1)=1+1$

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

$\therefore \Sigma \frac{p}{p-a}=2$