Question

Q.15. If x, y, z are positive numbers, then value of the determinant $\left|\begin{array}{ccc}1& {\log }_{x}y& {\log }_{x}z\\ {\log }_{y}x& 1& {\log }_{y}z\\ {\log }_{z}x& {\log }_{z}y& 1\end{array}\right|$ is equal to -
(1) 0
(2) 3
(3) log xyz
(4) none

Answer 1

$$\begin{gathered} \left|\begin{array}{ccc}1& {\log }_x\left(y\right)& {\log }_x\left(z\right)\\ {\log }_y\left(x\right)& 1& {\log }_y\left(z\right)\\ {\log }_z\left(x\right)& {\log }_z\left(y\right)& 1\end{array}\right| \\ Expandthedeterminant \\ 1(1-{\log }_y\left(z\right).{\log }_z\left(y\right))-{\log }_x\left(y\right)({\log }_y\left(x\right)-{\log }_y\left(z\right).{\log }_z\left(x\right))+{\log }_x\left(z\right)({\log }_y\left(x\right).{\log }_z\left(y\right)-{\log }_z\left(x\right)) \\ \because {\log }_a\left(b\right)=\frac{\log \left(b\right)}{\log \left(a\right)} \\ 1-\frac{\log z}{\log y}.\frac{\log y}{\log z}-\frac{\log y}{\log x}(\frac{\log x}{\log y}-\frac{\log z}{\log y}.\frac{\log x}{\log z})+\frac{\log z}{\log x}(\frac{\log x}{\log y}.\frac{\log y}{\log z}-\frac{\log x}{\log z}) \\ =0 \end{gathered}$$