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If x,y,zx,y,z are non-zero real numbers such that ∣10+x10101010+y10101010+z∣=0\begin{vmatrix} 10+x & 10 & 10 \\ 10 & 10+y & 10 \\ 10 & 10 & 10+z \end{vmatrix}=0, then the value of x−1+y−1+z−1x^{-1}+y^{-1}+z^{-1} is

Solution

✅ Correct Option: 4

Expanding the determinant gives xyz+10(xy+yz+zx)=0xyz + 10(xy+yz+zx) = 0.

Dividing throughout by xyzxyz, which is non-zero:

1+10(1x+1y+1z)=01 + 10\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right) = 0.

Hence x−1+y−1+z−1=−110x^{-1}+y^{-1}+z^{-1} = -\dfrac{1}{10}.

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