For positive numbers $x$, $y$ and $z$, the numerical value of the determinant
\begin{vmatrix}
1 & \log_xy & \log_xz \\
\log_yx & 1 & \log_yz \\
\log_zx & \log_zy & 1 \\
\end{vmatrix}
is
(A) $0$
(B) $\log \ xyz$
(C) $\log \ (x+y+z)$
(D) $\log \ x \ \log \ y \ \log \ z$
My solution:
I changed the bases of the elements. I wrote $\log_yx$ as $\frac{\log x}{\log y}$. I did the same for the others. Then, I took $\log x$, $\log y$ and $\log z$ common from columns one, two and three respectively. I then multiplied $\log x$, $\log y$ and $\log z$ to rows one, two and three respectively. I got the following determinant:
\begin{vmatrix}
\log x & 1 & 1 \\
1 & \log y & 1 \\
1 & 1 & \log z \\
\end{vmatrix}
I used cofactor expansion to get the value of the determinant as
$\log \ x \ \log \ y \ \log \ z – \log \ xyz + 2$
What am I doing wrong, and what is the correct solution?
Thank you so much!
Best Answer
Multiplying the first, second and third rows by $\log{x}$, $\log{y}$ and $\log{z}$ respectively gives $$\begin{vmatrix} \log{x} & \log{y} & \log{z} \\ \log{x} & \log{y} & \log{z} \\ \log{x} & \log{y} & \log{z} \\ \end{vmatrix}=0$$ If you calculate the determinant normally you get $$\begin{vmatrix} 1 & \log_xy & \log_xz \\ \log_yx & 1 & \log_yz \\ \log_zx & \log_zy & 1 \\ \end{vmatrix}$$ $$=1+\log_y{x}\log_z{y}\log_x{z}+\log_z{x}\log_x{y}\log_y{z}-\log_x{z}\log_z{x}-\log_y{z}\log_z{y}-\log_x{y}\log_y{x}$$ $$=1+\frac{\ln{x}\ln{y}\ln{z}}{\ln{y}\ln{z}\ln{x}}+\frac{\ln{x}\ln{y}\ln{z}}{\ln{z}\ln{x}\ln{y}}-\frac{\ln{z}\ln{x}}{\ln{x}\ln{z}}-\frac{\ln{z}\ln{y}}{\ln{y}\ln{z}}-\frac{\ln{y}\ln{x}}{\ln{x}\ln{y}}$$ $$=1+1+1-1-1-1=0$$