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 $\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ is equal to _____.

Answer 1

Answer: -3

$\Delta =\left|\begin{array}{ccc}1+x& 1& 1\\ 1+y& 1+2y& 1\\ 1+z& 1+z& 1+3z\end{array}\right|=0$
Applying $C_1\to C_1-C_3,C_2\to C_2-C_3$ we get
$\Delta =\left|\begin{array}{ccc}x& 0& 1\\ y& 2y& 1\\ -2z& -2z& 1+3z\end{array}\right|=0$
$\therefore [2xy+6zxy+4yz+2zx-2yz]=0$
$\therefore 2(xyz)\left[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3\right]=0$
$\therefore \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=-3$