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 $-1$

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

$\Rightarrow \left|\begin{array}{ccc}x& -y& 0\\ 0& y& -z\\ 1& 1& 1+z\end{array}\right|=0$, use $R_1\to R_1-R_2$ and $R_2\to R_2-R_3$

Now expand along first row,

$\Rightarrow x\left[y\right(1+z)+z]+y(0+z)=0$

$\Rightarrow xy(1+z)+xz+yz=0$

$\Rightarrow xy+yz+zx=-xyz$

$\Rightarrow x^{-1}+y^{-1}+z^{-1}=-1$, divide both sides by $xyz$