Question

If $A=\left[\begin{array}{ccc}1& {\log }_{x}y& {\log }_{x}z\\ {\log }_{y}x& 4& {\log }_{y}z\\ {\log }_{z}x& {\log }_{z}y& 6\end{array}\right]$ equals to,

where $\log x={\log }_{e}x$
$\log y={\log }_{e}y$
$\log z={\log }_{e}z$

• A. $15$
• B. $0$
• C. $-1$
• D. $1$

Answer 1

The correct option is A $15$

$A=\left[\begin{array}{ccc}1& {\log }_{x}y& {\log }_{x}z\\ {\log }_{y}x& 4& {\log }_{y}z\\ {\log }_{z}x& {\log }_{z}y& 6\end{array}\right]$
$|A|=\left|\begin{array}{ccc}\frac{\log x}{\log x}& \frac{\log y}{\log x}& \frac{\log z}{\log x}\\ \frac{\log x}{\log y}& \frac{4\log y}{\log y}& \frac{\log z}{\log y}\\ \frac{\log x}{\log z}& \frac{\log y}{\log z}& \frac{6\log z}{\log z}\end{array}\right|$
Taking out common $\frac{1}{\log x},\frac{1}{\log y},\frac{1}{\log z}$ from $R_1,R_2,R_3$
$=\frac{1}{\log x\log y\log z}\left|\begin{array}{ccc}\log x& \log y& \log z\\ \log x& 4\log y& \log z\\ \log x& \log y& 6\log z\end{array}\right|$
$=\left|\begin{array}{ccc}1& 1& 1\\ 1& 4& 1\\ 1& 1& 6\end{array}\right|$
Applying $R_1\to R_1-R_3$
$=\left|\begin{array}{ccc}0& 0& -5\\ 1& 4& 1\\ 1& 1& 6\end{array}\right|$
$=-5(1-4)=15$