Question

If $x,y,z$ are non-zero real numbers and $\left|\begin{array}{ccc}1+x& 1& 1\\ 1+y& 1+2y& 1\\ 1+z& 1+z& 1+3z\end{array}\right|=0$ then $-(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$ is equal to________

Answer 1

$\left|\begin{array}{ccc}1+x& 1& 1\\ 1+y& 1+2y& 1\\ 1+z& 1+z& 1+3z\end{array}\right|dividing{R}_1,R_2\&R_{3}by{x}_{1}y\&zrepectively=xyz\left|\begin{array}{ccc}1+\frac{1}{x}& \frac{1}{x}& \frac{1}{x}\\ 1+\frac{1}{y}& 2+\frac{1}{y}& \frac{1}{y}\\ 1+\frac{1}{z}& 1+\frac{1}{x}& 3+\frac{1}{z}\end{array}\right|=0=xyz\left|\begin{array}{ccc}3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}& 3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}& 3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\\ 1+\frac{1}{y}& 2+\frac{1}{y}& \frac{1}{y}\\ 1+\frac{1}{z}& 1+\frac{1}{z}& 3+\frac{1}{z}\end{array}\right|=0Takingoutcommon(3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z})from{R}_1=\left(xyz\right)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3)\left|\begin{array}{ccc}1& 1& 1\\ 1+\frac{1}{y}& 2+\frac{1}{y}& \frac{1}{y}\\ 1+\frac{1}{z}& 1+\frac{1}{z}& 3+\frac{1}{z}\end{array}\right|=0=\left(xyz\right)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3)\left|\begin{array}{ccc}1& 0& 0\\ 1+\frac{1}{y}& 1& -1\\ 1+\frac{1}{z}& 0& 2\end{array}\right|=0Expandingalong{R}_1$

$\begin{array}{c}\Rightarrow xy=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3\right)\left[2-0\right]=0\\ \Rightarrow 2xy=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3\right)=0\\ \Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3=0\\ \Rightarrow -\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3\end{array}$