For positive numbers $x, y$ and $z$, the numerical value of the determinant $\left|\begin{array}{ccc

Question

For positive numbers $x, y$ and $z$, the numerical value of the determinant $\left|\begin{array}{ccc}1 & \log _{x} y & \log _{x} z \ \log _{y} x & 1 & \log _{y} z \ \log _{z} x & \log _{z} y & 1\end{array}ight|$ is ..................

Professor PC Thomas Classes · Answer

Given determinant $ \begin{aligned} & =1\left(1-\log _{y} z \log _{z} yight)-\log _{y} x\left(\log _{x} y-\log _{x} zight. \ & \left.\log _{z} yight)+\log _{z} x\left(\log _{y} z \log _{x} y-\log _{x} zight) \ & =1(0)-\log _{y} x(0)+\log _{z} x(0) \end{aligned} $ $ \left(\because \log _{y} x=\frac{\log x}{\log y}ight) $ $=0$ $$ =\left|\begin{array}{lll} 1 & a & a^{2} \ 1 & b & b^{2} \ 1 & c & c^{2} \end{array}ight|-\left|\begin{array}{lll} 1 & a & b c \ 1 & b & a c \ 1 & c & a b \end{array}ight| $$ $=\left|\begin{array}{lll}1 & a & a^{2} \ 1 & b & b^{2} \ 1 & c & c^{2}\end{array}ight|-\frac{1}{a b c}\left|\begin{array}{lll}a & a^{2} & a b c \ b & b^{2} & a b c \ c & c^{2} & a b c\end{array}ight|$\ $=\left|\begin{array}{lll}1 & a & a^{2} \ 1 & b & b^{2} \ 1 & c & c^{2}\end{array}ight|-\left|\begin{array}{lll}1 & a & a^{2} \ 1 & b & b^{2} \ 1 & c & c^{2}\end{array}ight|=0$\