Q101 / 180 · KCET 2009 · Mathematics

The value of $\left|\begin{array}{ccc}1 & \log _{\mathrm{x}} \mathrm{y} & \log _{\mathrm{x}} \mathrm{z} \\ \log _{\mathrm{y}} \mathrm{x} & 1 & \log _{\mathrm{y}} \mathrm{z} \\ \log _{\mathrm{z}} \mathrm{x} & \log _{\mathrm{z}} \mathrm{y} & 1\end{array}\right|$ is equal to

  • Correct answer
    A
    0
  • B
    1
  • C
    $x y z$
  • D
    $\log x y z$

Solution

$\left|\begin{array}{ccc}1 & \log _{\mathrm{x}} \mathrm{y} & \log _{\mathrm{x}} \mathrm{z} \\ \log _{\mathrm{y} x} & 1 & \log _{\mathrm{y} z} \\ \log _{\mathrm{z}} \mathrm{x} & \log _{\mathrm{z}} \mathrm{y} & 1\end{array}\right|$ $=1\left(1-\log _{\mathrm{y}} \mathrm{z} \log _{\mathrm{z}} \mathrm{y}\right)$ $-\log _{\mathrm{x}} \mathrm{y}\left(\log _{\mathrm{y}} \mathrm{x}-\log _{\mathrm{z}} \mathrm{x} \log _{\mathrm{y}} \mathrm{z}\right)$ $\quad+\log _{\mathrm{x}} \mathrm{z}\left(\log _{\mathrm{z}} \mathrm{y}\left(\log _{\mathrm{y}} \mathrm{x}-\log _{\mathrm{z}} \mathrm{x}\right)\right.$ $=\left(1-\log _{\mathrm{y}} \mathrm{y}\right)-\log _{\mathrm{x} y}\left(\log _{\mathrm{y}} \mathrm{x}-\log _{\mathrm{y}} \mathrm{x}\right)$ $+\log _{\mathrm{x}} \mathrm{z}\left(\log _{\mathrm{z}} \mathrm{x}-\log _{\mathrm{z}} \mathrm{x}\right)$ $=(1-1)-0+0=0$

Hence, the correct answer is option: 'A'.

Relevant Questions