Question

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

Answer 1

$\Delta=\begin{vmatrix}1+x&1&1\\ 1+y&1+2y&1\\ 1+z&1+z&1+3z\end{vmatrix}=0$
Applying $C_{1} \rightarrow C_{1}-C_{3}, C_{2} \rightarrow C_{2}-C_{3}$ we get
$\Delta=\begin{vmatrix}x&0&1\\ y&2y&1\\ -2z&-2z&1+3z\end{vmatrix}=0$
$\therefore [2 x y+6 z x y+4 y z+2 z x-2 y z]=0$
$\therefore 2(x y z)\left[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3\right]=0$
$\therefore \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=-3$