Question

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

• A. $0$
• B. $1$
• C. ${\log }_{e}xyz$
• D. $-{\log }_{e}xyz$

Answer 1

The correct option is A $0$

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

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

Since, ${\log }_{a}b=\frac{\log a}{\log b}$

So simplifying above further we get

${\log }_{y}z\times {\log }_{z}y=\frac{\log z}{\log y}\times \frac{\log y}{\log z}=1$

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

$\Rightarrow D=0$