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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 to1. -12. - x - y - z3. x-1y-1z-14. xyz |
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Answer» 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 ⇒ \(\rm {1\over z}+ 1 +{1\over y}+ {1\over x} = 0\) ⇒ \(\boldsymbol{\rm x^{-1}+ y^{-1}+ z^{-1} = -1}\) |
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