Question

$\left|\begin{array}{ccc}1& {\log }_{x}y& {\log }_{x}z\\ {\log }_{y}x& 1& {\log }_{y}z\\ {\log }_{z}x& {\log }_{z}y& 1\end{array}\right|$ = 0

• A. True
• B. False

Answer 1

The correct option is A True

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

We know that ${\log }_{x}x={\log }_{y}y={\log }_{z}z=1$

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

$C_1\to C_1+C_2+C_3$

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

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

$C_1\to \frac{C_1}{{\log }_{x}xyz}$

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

Using ${\log }_{x}x={\log }_{y}y={\log }_{z}z=1$

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

$={\log }_{x}xyz[1(1-1)-{\log }_{x}y(1-{\log }_{y}z)+{\log }_{x}z({\log }_{z}y-1)]$

$={\log }_{x}xyz[0-{\log }_{x}y+{\log }_{x}y{\log }_{y}z+{\log }_{x}z{\log }_{z}y-{\log }_{x}z]$

$={\log }_{x}xyz[-{\log }_{x}y+{\log }_{x}z+{\log }_{x}y-{\log }_{x}z]$

$={\log }_{x}xyz\times 0=0$

Hence the given statement is true.