Mathematics - Determinant Question with Solution | TestHub

As the given system of equations has a non-trivial solution. $\Delta =\left|\begin{array}{lll}p& a& a\\ b& q& b\\ c& c& r\end{array}\right|=0$ Applying $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$ $\Delta =\left|\begin{array}{ccc}p& a-p& a-p\\ b& q-b& 0\\ c& 0& r-c\end{array}\right|=0$ Expanding along $C_3$, we get $(a-p)\left|\begin{array}{cc}b& q-b\\ c& 0\end{array}\right|+(r-c)\left|\begin{array}{cc}p& a-p\\ b& q-b\end{array}\right|=0$ $\begin{array}{rl}& \Rightarrow (a-p)(-c)(q-b)+(r-c)\\ & \{p(q-b)-b(a-p)\}=0\\ & \Rightarrow \\ & (p-a)(q-b)c+p(r-c)(q-b)+b(r-c)(p-a)=0\end{array}$ Dividing by $(p-a)(q-b)(r-c)$, we get $\frac{c}{r-c}+\frac{p}{p-a}+\frac{b}{q-b}=0$ $\Rightarrow \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=\frac{q-b}{q-b}+\frac{r-c}{r-c}=2$