Determinant

Algebra Level 3

If $x$ , $y$ and $z$ are positive real numbers, find the value of the determinant below:

$\large \begin{vmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 1 & \log_y z \\ \log_z x & \log_z y & 1 \end{vmatrix}$

\log_e xyz

-\log xyz

2 solutions

Chew-Seong Cheong
Nov 2, 2016

Relevant wiki: 3 by 3 Determinant

$\begin{aligned} \Delta & = \begin{vmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 1 & \log_y z \\ \log_z x & \log_z y & 1 \end{vmatrix} \\ & = \begin{vmatrix} 1 & \frac {\log y}{\log x} & \frac {\log z}{\log x} \\ \frac {\log x}{\log y} & 1 & \frac {\log z}{\log y} \\ \frac {\log x}{\log z} & \frac {\log y}{\log z} & 1 \end{vmatrix} \\ & = \small 1 \times 1 \times 1 + \frac {\log y}{\log x} \times \frac {\log z}{\log y} \times \frac {\log x}{\log z} + \frac {\log z}{\log x} \times \frac {\log x}{\log y} \times \frac {\log y}{\log z} \\ & \quad \small - 1 \times \frac {\log z}{\log y} \times \frac {\log y}{\log z} - \frac {\log y}{\log x} \times \frac {\log x}{\log y} \times 1 - \frac {\log z}{\log x} \times 1 \times \frac {\log x}{\log z} \\ & = 1+1+1-1-1-1 \\ & = \boxed{0} \end{aligned}$

Other than expanding out the determinant, is there another approach that we can take?

Calvin Lin Staff - 4 years, 7 months ago

Calvin Lin Staff
Nov 16, 2016

Consider the first 2 columns of the matrix. We have

$\log_x y \begin{pmatrix} 1 \\ \log_y x \\ \log_z x \\ \end{pmatrix} = \begin{pmatrix} \log_x y \\ \frac{ \log y}{\log x} \times \frac{ \log x} { \log y } \\ \frac{ \log y}{\log x} \times \frac{ \log x} { \log z } \\ \end{pmatrix} = \begin{pmatrix} \log_x y \\ 1 \\ \log_z x \\ \end{pmatrix}.$

Since the second column is a linear multiple of the first column, the determinant of the matrix is 0.

Determinant

2 solutions

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