Question

If $xyz$ are all different and not equal to zero and $M = \begin{vmatrix} 1+x &\, 1&\, 1 \\[0.3em] 1&1+y&1 \\[0.3em] 1&1&1+z \end{vmatrix} = 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 $\begin{vmatrix}1+x&1&1\\ -x&y&0\\ 0&-y&z\end{vmatrix} = 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+x y+y z+x z+x y z=x+y+z+1 $
$\Rightarrow x y+y z+x z=-x y z$
On dividing both sides by $x y z$, we get
$\frac{1}{z}+\frac{1}{x}+\frac{1}{y}=-1 $
$\Rightarrow x^{-1}+y^{-1}+z^{-1}=-1$