ISI2016-DCG-22

Question

The value of $\:\:\begin{vmatrix} 1&\log_{x}y &\log_{x}z \\ \log_{y}x &1 &\log_{y}z \\\log_{z}x & \log_{z}y&1 \end{vmatrix}\:\:$ is

• A. $0$
• B. $1$
• C. $-1$
• D. None of these

Comments

put x=y=z=k(some constant). This will make all the rows same.Hence determinant will be 0.

Answer 1

Answer: $\mathbf A$

Explanation:

We know that:

$\color {blue} {\mathbf {\log_ab = \frac{\log b}{\log a}}}$

$\therefore$ On solving the determinant we, get:

$ \mathrm {1(1-\log_zy\log_yz) -\log_xy(1.\log_yx-\log_zx\log_xz) + \log_xz(1.\log_zx-\log_yx\log_zy)}$

$= \mathrm {1\left (1-\frac{{\log y}}{\log z}\frac{\log z}{\log y}\right ) - \frac{\log y}{\log x}\left (\frac{\log x}{\log y}-\frac{\log x}{\log z}\frac{\log z}{\log x}\right ) + \frac{\log z}{\log x}\left ( \frac{\log x}{\log z}-\frac{\log x}{\log y}\frac{\log y}{\log z}\right) }$

$=0$

$\therefore \mathbf A$ is the correct option.