Question

If $x, y, z$ are all positive and are the $p^{th}, q^{th}$ and $r^{th}$ terms of a geometric progression respectively, then the value of the determinant $\begin{vmatrix} \log x & p & 1 \\\ \log y & q & 1 \\\ \log z & r & 1 \end{vmatrix} = 0$ equals

• A. $\log \, xyz$
• B. $(p -1) (q - 1)(r -1)$
• C. $pqr$
• D. 0

Answer 1

Answer: (D)

0

Answer 2

Let $a$ and $R$ be the first term and common ratio of a GP.
$\therefore T_p = aR^{P-1} = x$
$T_q = aR^{q-1} = y$
And $T_r = aR^{r-1} = z$
$\Rightarrow \ \log x = \log a + (p - 1) \log R$
$\ \log y = \log a + (q - 1) \log R$
and $\log z = \log a + (r -1) \log R$
$\therefore \begin{vmatrix} \log x & x & 1 \\\ \log y & y & 1 \\\ \log z & z & 1 \end{vmatrix}= \begin{vmatrix} \log a + p - 1 & \log R & p & 1 \\\ \log a + q - 1 & \log R & q & 1 \\\ \log a + r - 1 & \log R & r & 1 \end{vmatrix}$
$= \begin{vmatrix} \log a & p & 1 \\\ \log a & q & 1 \\\ \log a & r & 1 \end{vmatrix} + \begin{vmatrix} p - 1 & \log R & p & 1 \\\ q - 1 & \log R & q & 1 \\\ r - 1 & \log R & r & 1 \end{vmatrix}$
$= \log a \ \ \begin{vmatrix} 1 & p & 1 \\\ 1 & q & 1 \\\ 1& r & 1 \end{vmatrix} + \log R \ \ \begin{vmatrix} p - 1 & p - 1 & 1 \\\ q - 1 & q - 1 & 1 \\\ r - 1 & r - 1 & 1 \end{vmatrix}$
$C_2 \to C_2 - C_3$
$= 0 + 0 = 0$ ($\because$ two columns are identical)