If `x , y , z` are different from zero and `|1+x1 1 1 1+y1 1 1 1+z|=0` then the value of `x^(-1)+y^(-1)+2^(-1)` is `x y z` (b) `x^(-1)y^(-1)z^(-1)` (c) `-x-y-z` (d) `-1`

Options

A. `xyz`
B. `x^(-1)y^(-1)z^(-1)`
C. `-x-y-z`
D. `-1`

Answer

D

Explanation

We have `|(1+x,1,1),(1,1+y,1),(1,1,1+z)|=0`
Applying `C_(1)toC_(1)-C_(3)` and `C_(2)toC_(2)-C_(3)`
`implies |(x,0,1),(0,y,1),(-z,-z,1+z)|=0`
Expanding along `R_(1)`
`x[y(a+z)+z]-0+1(yz)=0`
`impliesx(y+yz+z)+yz=0`
`implies xy+xyz+xz+yz=0`
`= (xy)/(x y)=(x yz)/(x yz)+(xz)/(x yz)+(yz)/(x yz)=0` [on dividing `(xyz)` from both sides]
`implies 1/x+1/y+1/z+1=0`
`implies 1/x+1/y+1/z=-1`
`:. x^(-1)+y^(-1)+z^(-1)=-1`