For positive numbers x- y and z the numerical value of the determinant Nbeginvmatrix 1 -log - x y -log - x z log - y x - 1 -log - y z log - z x -log - z y - 1lend vmatrix isA- log e x y zB- -log x y zC- 1D- 0

The correct option is D 01 amp; log_xy amp; log_x z log_y x amp; 1 amp; log_y z log_z x amp; log_z y amp; 1 =(1 log_zy log_yz) log_xy ...

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Standard XII

Mathematics

Determinant

For positive ...

Question

For positive numbers x, y and z the numerical value of the determinant $∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & l o g_{x} y & l o g_{x} z l o g_{y} x & 1 & l o g_{y} z l o g_{z} x & l o g_{z} y & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$ is

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l o g_{e} x y z

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-log xyz

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Solution

$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & l o g_{x} y & l o g_{x} z l o g_{y} x & 1 & l o g_{y} z l o g_{z} x & l o g_{z} y & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = (1 - l o g_{z} y l o g_{y} z) - l o g_{x} y (l o g_{y} x - l o g_{z} x l o g_{y} z) + l o g_{x} z (l o g_{y} x l o g_{z} y - l o g_{z} x) = (1 - 1) - (1 - l o g_{x} y l o g_{y} x) + (l o g_{x} z l o g_{z} x - 1) = 0 {S i n c e l o g_{x} y . l o g_{y} x = 1} .$

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For positive numbers x, y and z the numerical value of the determinant ∣∣ ∣ ∣∣1logxylogxzlogyx1logyzlogzxlogzy1∣∣ ∣ ∣∣ is

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