Question

If $x,y,z$ are all different and not equal to zero and $\left|\begin{array}{ccc}1+x& 1& 1\\ 1& 1+y& 1\\ 1& 1& 1+z\end{array}\right|=0$ then the value of $x^{-1}+y^{-1}+z^{-1}$ is equal to

• A. $xyz$
• B. $x^{-1}{y}^{-1}{z}^{-1}$
• C. -x-y-z
• D. $-1$

Answer 1

The correct option is D $-1$

$\left|\begin{array}{ccc}1+x& 1& 1\\ 1& 1+y& 1\\ 1& 1& 1+z\end{array}\right|=0$

$xyz\left|\begin{array}{ccc}\frac{1}{x}+1& \frac{1}{x}& \frac{1}{x}\\ \frac{1}{y}& \frac{1}{y}+1& \frac{1}{y}\\ \frac{1}{z}& \frac{1}{z}& \frac{1}{z}+1\end{array}\right|=0$

$R_1\to R_1+R_2+R_3$
$xyz\left|\begin{array}{ccc}1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}& 1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}& 1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\\ \frac{1}{y}& \frac{1}{y}+1& \frac{1}{y}\\ \frac{1}{z}& \frac{1}{z}& \frac{1}{z}+1\end{array}\right|=0$

$xyz(1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\left|\begin{array}{ccc}1& 1& 1\\ \frac{1}{y}& \frac{1}{y}+1& \frac{1}{y}\\ \frac{1}{z}& \frac{1}{z}& \frac{1}{z}+1\end{array}\right|=0$

$C_2\to C_2-C_1,C_3\to C_3-C_1$
$xyz(1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\left|\begin{array}{ccc}1& 0& 0\\ \frac{1}{y}& 1& 0\\ \frac{1}{z}& 0& 1\end{array}\right|=0$

$xyz(1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\left[1\right(1-0\left)\right]=0$
$1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$
$x^{-1}+y^{-1}+z^{-1}=-1$