Question

If $x,y$ and $z$ be greater than $1$, then the value of $\left|\begin{array}{ccc}1& {\log }_{x}y& {\log }_{x}z\\ {\log }_{y}x& 1& {\log }_{y}z\\ {\log }_{z}x& {\log }_{z}y& 1\end{array}\right|$ is

• A. $\log x.\log y.\log z$
• B. $\log x+\log y+\log z$
• C. $0$
• D. $\frac{1}{\left(\log x\right).\left(\log y\right).\left(\log z\right)}$

Answer 1

The correct option is C $0$

$\left|\begin{array}{ccc}1& {\log }_{x}y& {\log }_{x}z\\ {\log }_{y}x& 1& {\log }_{y}z\\ {\log }_{z}x& {\log }_{z}y& 1\end{array}\right|$

Expand along first row

$=1(1-{\log }_{z}y\cdot {\log }_{y}z)-{\log }_{x}y({\log }_{y}x-{\log }_{z}x\cdot {\log }_{y}z+{\log }_{x}z({\log }_{y}x\cdot {\log }_{z}y-{\log }_{z}x)$

$=1(1-1)-{\log }_{x}y({\log }_{y}x-{\log }_{y}x)+{\log }_{x}z({\log }_{z}x-{\log }_{z}x)$, use base change formula ${\log }_{a}b=\frac{\log b}{\log a}$

$=0$