InterviewSolution
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InterviewSolution
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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`A. `xyz`B. `x^(-1)y^(-1)z^(-1)`C. `-x-y-z`D. `-1` |
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Answer» Correct Answer - D 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` |
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