The numerical value of this determinant is?

Question

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!

Answer 1

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 $$\begin{vmatrix} 1 & \log_xy & \log_xz \\ \log_yx & 1 & \log_yz \\ \log_zx & \log_zy & 1 \\ \end{vmatrix}$$ $$=1+\log_y{x}\log_z{y}\log_x{z}+\log_z{x}\log_x{y}\log_y{z}-\log_x{z}\log_z{x}-\log_y{z}\log_z{y}-\log_x{y}\log_y{x}$$ $$=1+\frac{\ln{x}\ln{y}\ln{z}}{\ln{y}\ln{z}\ln{x}}+\frac{\ln{x}\ln{y}\ln{z}}{\ln{z}\ln{x}\ln{y}}-\frac{\ln{z}\ln{x}}{\ln{x}\ln{z}}-\frac{\ln{z}\ln{y}}{\ln{y}\ln{z}}-\frac{\ln{y}\ln{x}}{\ln{x}\ln{y}}$$ $$=1+1+1-1-1-1=0$$

Answer 2

Let me follow your thought:

Taking $\log x, \log y, \log z$ out of the columns.

$$\log x \log y \log z \begin{vmatrix} \frac1{\log x} & \frac1{\log x} & \frac1{\log x} \\ \frac1{\log y} & \frac1{\log y} & \frac1{\log y} \\ \frac1{\log z} & \frac1{\log z} & \frac1{\log z} \end{vmatrix}$$

You then multiply $\log x, \log y, \log z$ to the rows. This would give you the determinant of the all ones matrix, hence the determinant is $0$.