Question

If $\left|\begin{array}{ccc}a& b-y& c-z\\ a-x& b& c-z\\ a-x& b-y& c\end{array}\right|=$0, then using properties of determinants, find the value of $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}$, where $x,y,z\ne$0.

Answer 1

$\left|\begin{array}{ccc}a& b-y& c-z\\ a-x& b& c-z\\ a-x& b-y& c\end{array}\right|=$0

$$\begin{gathered} R_1\to R_1-R_2 \\ \left|\begin{array}{ccc}x& -y& 0\\ a-x& b& c-z\\ a-x& b-y& c\end{array}\right|=0 \\ {R}_2\to R_2-R_3 \\ \Rightarrow \left|\begin{array}{ccc}x& -y& 0\\ 0& y& -z\\ a-x& b-y& c\end{array}\right|=0 \\ \mathrm{Expanding}\mathrm{along}\mathrm{first}\mathrm{row},\mathrm{we}\mathrm{get} \\ x(yc+zb-zy)+y(0-za+zx)=0 \\ \Rightarrow xyc+xzb-xyz+zya-xyz=0 \\ xyc+xzb-2xyz+zya=0 \\ \mathrm{Dividing}\mathrm{by}\mathrm{xyz},\mathrm{we}\mathrm{get} \\ \frac{c}{z}+\frac{b}{y}-2+\frac{a}{x}=0 \\ \therefore \frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2 \end{gathered}$$