Question

If x, y, z are all positive and are the pth, qth and rth terms of a geometric progression respectively, then find the value of the determinant \(\begin{bmatrix}\text{log}\,\,x&p&1\\\text{log}\,\,y&q&1\\\text{log}\,\,z&r&1\end{bmatrix}\)

Answer 1

Let a be the first term and R the common ratio of the G.P.

Then, T_p = a\(R\)^{p – 1} = x 

T_q = a\(R\)^{q – 1} = y 

T_r = a\(R\)^{r – 1} = z 

⇒ log \(x\) = log (a\(R\)^{p – 1}) = log a + (p – 1) log \(R\) 

log y = log (a\(R\)^{q – 1}) = log a + (q – 1) log \(R\) 

log z = log (a\(R\)^{r – 1}) = log a + (r – 1) log \(R\)

∴ \(\begin{bmatrix}\text{log}\,\,x&p&1\\\text{log}\,\,x&q&1\\\text{log}\,\,z&r&1\end{bmatrix}\) = \(\begin{bmatrix}\text{log}\,\,a\,+(p-1)\,\text{log}\,R&p&1\\\text{log}\,\,a\,+(q-1)\,\text{log}\,R&q&1\\\text{log}\,\,a\,+(r-1)\text{log}\,R&r&1\end{bmatrix}\)

= \(\begin{bmatrix}\text{log}\,\,a&p&1\\\text{log}\,\,a&q&1\\\text{log}\,\,a&r&1\end{bmatrix}\)+ \(\begin{bmatrix}(p-1)\,\text{log}\,R&p&1\\(q-1)\,\text{log}\,R&q&1\\(r-1)\text{log}\,R&r&1\end{bmatrix}\)

= log a \(\begin{bmatrix}1&p&1\\1&q&1\\1&r&1\end{bmatrix}\) + log \(R\) \(\begin{bmatrix}(p-1)&p&1\\(q-1)&q&1\\(r-1)&r&1\end{bmatrix}\)

=log a x 0 + log \(R\) \(\begin{bmatrix}1&p&1\\1&q&1\\1&r&1\end{bmatrix}\) - log \(R\) \(\begin{bmatrix}1&p&1\\1&q&1\\1&r&1\end{bmatrix}\)

= 0 + log \(R\) x 0 - log \(R\) x 0 = 0        

(∵ The value of a determinants with two identical rows or columns is zero).