Question

If x y z are all different and not equal to zero and \(\left| {\begin{array}{*{20}{c}} {1 + x}&1&1\\ 1&{1 + y}&1\\ 1&1&{1 + z} \end{array}} \right| = 0\) then the value of x^-1 + y^-1 + z^-1 is equal to
1. -1
2. - x - y - z
3.  x-1y-1z-1
4. xyz

Answer 1

Correct Answer - Option 1 : -1

Calculation:

\(\left| {\begin{array}{*{20}{c}} {1 + x}&1&1\\ 1&{1 + y}&1\\ 1&1&{1 + z} \end{array}} \right| = 0\)

R₁ = R₁ - R₂

\(\left| {\begin{array}{*{20}{c}} {x}&-y&0\\ 1&{1 + y}&1\\ 1&1&{1 + z} \end{array}} \right| = 0\)

R₂ = R₂ - R₃

\(\left| {\begin{array}{*{20}{c}} {x}&-y&0\\ 0&{y}&-z\\ 1&1&{1 + z} \end{array}} \right| = 0\)

Now, Expanding along R₁, we get

⇒ x [y(1 + z) - (-z)] - (-y) [0 - (-z)] + 0 = 0

⇒ x [y + yz + z] + y [z] = 0

⇒ xy + xyz + xz + yz = 0

Dividing by xyz

⇒ \(\rm {1\over z}+ 1 +{1\over y}+ {1\over x} = 0\)

⇒ \(\boldsymbol{\rm x^{-1}+ y^{-1}+ z^{-1} = -1}\)