For positive numbers $x,\ y\ and\mathcal{\ z,\ }$the numerical value of the determinant $\left| \begin{matrix} 1 & \log_{x}y & \log_{x}\mathcal{z} \ \log_{y}x & 1 & \log_{y}\mathcal{z} \ \log_{\mathcal{z}}x & \log_{\mathcal{z}}y & 1 \end{matrix} ight|$ is

Question

For positive numbers $x,\ y\ and\mathcal{\ z,\ }$the numerical value of the determinant $\left| \begin{matrix} 1 & \log_{x}y & \log_{x}\mathcal{z} \ \log_{y}x & 1 & \log_{y}\mathcal{z} \ \log_{\mathcal{z}}x & \log_{\mathcal{z}}y & 1 \end{matrix} ight|$ is

Dr. Abubacker · Accepted Answer

(a)
Let,$\ \mathrm{\Delta} = \left| \begin{matrix}
1 & \log_{x}y & \log_{x}\mathcal{z} \
\log_{y}x & 1 & \log_{y}\mathcal{z} \
\log_{\mathcal{z}}x & \log_{\mathcal{z}}y & 1
\end{matrix} ight|$
$= 1\left( 1 - \log_{\mathcal{z}}y\ \log_{y}\mathcal{z} ight) - \log_{x}y\left( \log_{y}x - \log_{y}\mathcal{z}\ \log_{\mathcal{z}}x ight) + \log_{x}\mathcal{z}(\log_{y}x\ \log_{\mathcal{z}}y - \log_{\mathcal{z}}x$
$= (1 - \log_{\mathcal{z}}\mathcal{z) -}\log_{x}y\left( \log_{y}x - \log_{y}\mathcal{z}\ \log_{\mathcal{z}}x ight) + \log_{x}\mathcal{z}(\log_{y}{x\log_{\mathcal{z}}y} - \log_{\mathcal{z}}x)$
$= (1 - 1) - \left( 1 - \log_{x}y\log_{y}x ight) + \left( \log_{x}\mathcal{z}\log_{\mathcal{z}}x - 1 ight) = 0\$ (Since,$\ \log_{x}y\ \log_{y}x = 1)$
$= 0 - (1 - 1) + (1 - 1) = 0$

Q. For positive numbers $x,\ y\ and\mathcal{\ z,\ }$the numerical value of the determinant $\left| \begin{matrix} 1 & \log_{x}y & \log_{x}\mathcal{z} \\ \log_{y}x & 1 & \log_{y}\mathcal{z} \\ \log_{\mathcal{z}}x & \log_{\mathcal{z}}y & 1 \end{matrix} \right|$ is

Explanation

Reviewed by Dr. Abubacker

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