Q62 / 125 · BITSAT 2009 · Mathematics

If \(\mathrm{a} > 0, \mathrm{~b} > 0, \mathrm{c} > 0\) are respectively the \(\mathrm{pth}, \mathrm{q}\) th, rth terms of GP., then the value of the determinant \(\left|\begin{array}{lll}\log \mathrm{a} & \mathrm{p} & 1 \\ \log \mathrm{b} & \mathrm{q} & 1 \\ \log \mathrm{c} & \mathrm{r} & 1\end{array}\right|\) is

  • Correct answer
    A
    0
  • B
    1
  • C
    -1
  • D
    None of these

Solution

Let \(A\) be the 1 st term and \(\mathrm{R}\) the common ratio of G.P., then \(\begin{aligned} & a=T_p=A R^{p-1} \\ & \therefore \log \mathrm{a}=\log \mathrm{A}+(\mathrm{p}-1) \log \mathrm{R} \end{aligned}\) Similarly, \(\log \mathrm{b}=\log \mathrm{A}+(\mathrm{q}-1) \log \mathrm{R}\) and \(\log \mathrm{c}=\log \mathrm{A}+(\mathrm{r}-1) \log \mathrm{R}\) \(\therefore \Delta=\left|\begin{array}{lll} \log A+(p-1) \log R & p & 1 \\ \log A+(q-1) \log R & q & 1 \\ \log A+(r-1) \log R & r & 1 \end{array}\right|\) Split into two determinants and in the first take \(\log A\) common and in the second take \(\log \mathrm{R}\) common \(\Delta=\log A\left|\begin{array}{lll} 1 & p & 1 \\ 1 & q & 1 \\ 1 & r & 1 \end{array}\right|+\log R\left|\begin{array}{lll} p-1 & p & 1 \\ q-1 & q & 1 \\ r-1 & r & 1 \end{array}\right|\) Apply \(\mathrm{C}_1 \rightarrow \mathrm{C}_2 \rightarrow \mathrm{C}_3\) in the second \(\Delta=0+\log R\left|\begin{array}{lll} 0 & p & 1 \\ 0 & q & 1 \\ 0 & r & 1 \end{array}\right|=0\)

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

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