Question

If $xyz$ are all different and not equal to zero and $M=\left|\begin{array}{ccc}1+x& 1& 1\\ 1& 1+y& 1\\ 1& 1& 1+z\end{array}\right|=0$
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

Answer: (D)

Given $\left|\begin{array}{ccc}1+x& 1& 1\\ -x& y& 0\\ 0& -y& z\end{array}\right|=0$
Expanding along $R_1$, we get
$(1+x)[(1+y)(1+z)-1]-1(1+z-1)+1(1-1-y)=0$
$\Rightarrow (1+x)(1+y)(1+z)-(1+x)-z-y=0$
$\Rightarrow (1+x)(1+y)(1+z)=x+y+z+1$
$\Rightarrow 1+x+y+z+xy+yz+xz+xyz=x+y+z+1$
$\Rightarrow xy+yz+xz=-xyz$
On dividing both sides by $xyz$, we get
$\frac{1}{z}+\frac{1}{x}+\frac{1}{y}=-1$
$\Rightarrow x^{-1}+y^{-1}+z^{-1}=-1$