Question

If a, b, c are in G.P. with common ratio ${r_1}$ and $\alpha $ , $\beta $ , $\gamma $ also form a G.P. with common ratio ${r_2}$ , then find the conditions that ${r_1}$ and ${r_2}$ must satisfy so that the equations $ax + \alpha y + z = 0$ , $bx + \beta y + z = 0$ , $cx + \gamma y + z = 0$ have only zero solution.

Answer 1

Hint:
Since a, b, c are in G.P. with a common ratio ${r_1}$ , there is a relation between a, b and c. Similarly $\alpha $ ,$\beta $ , $\gamma $ also form a G.P. with a common ratio ${r_2}$ , hence, there is a relation between $\alpha $ , $\beta $ and $\gamma $ . We will use this relation in solving the determinant formed by the set of the equation that has zero solutions to get our answer.

Complete step by step solution:
Since a, b , c are in G.P. with a common ratio ${r_1}$ , therefore
⇒\[a \ne 0\] … (1)
⇒\[b = a{r_1}\] … (2)
⇒\[c = a{r_1}^2\] … (3)
Since$\alpha $ , $\beta $ , $\gamma $ are in G.P. with common ratio ${r_2}$ , therefore
⇒\[\alpha \ne 0\] … (4)
⇒\[\beta = \alpha {r_2}\] … (5)
⇒\[\gamma = \alpha {r_2}^2\] … (6)
Also according to the condition, the system of equations have only zero solution, hence,
⇒\[\left| {\begin{array}{*{20}{c}}
  a&\alpha &1 \\
  b&\beta &1 \\
  c&\gamma &1
\end{array}} \right| = 0\]
On substituting values from (2), (3), (5), and (6)
⇒\[\left| {\begin{array}{*{20}{c}}
  a&\alpha &1 \\
  {a{r_1}}&{\alpha {r_2}}&1 \\
  {a{r_1}^2}&{\alpha {r_2}^2}&1
\end{array}} \right| = 0\]
On taking a common from \[{C_1}\] and taking \[\alpha \] common from \[{C_2}\], we get
⇒\[a\alpha \left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  {{r_1}}&{{r_2}}&1 \\
  {{r_1}^2}&{{r_2}^2}&1
\end{array}} \right| = 0\]

On applying ${C_2} \to {C_2} - {C_1}$ and ${C_3} \to {C_3} - {C_1}$ , we get
⇒\[a\alpha \left| {\begin{array}{*{20}{c}}
  1&0&0 \\
  {{r_1}}&{{r_2} - {r_1}}&{1 - {r_1}} \\
  {{r_1}^2}&{{r_2}^2 - {r_1}^2}&{1 - {r_1}^2}
\end{array}} \right| = 0\]
We know that \[{a^2} - {b^2} = (a - b)(a + b)\] , so we have
⇒\[a\alpha \left| {\begin{array}{*{20}{c}}
  1&0&0 \\
  {{r_1}}&{{r_2} - {r_1}}&{1 - {r_1}} \\
  {{r_1}^2}&{({r_2} - {r_1})({r_2} + {r_1})}&{(1 + {r_1})(1 - {r_1})}
\end{array}} \right| = 0\]
On taking \[{r_2} - {r_1}\] common from \[{C_2}\] and taking \[1 - {r_1}\] common from \[{C_3}\], we get
⇒\[a\alpha ({r_2} - {r_1})(1 - {r_1})\left| {\begin{array}{*{20}{c}}
  1&0&0 \\
  {{r_1}}&1&1 \\
  {{r_1}^2}&{{r_2} + {r_1}}&{1 + {r_1}}
\end{array}} \right| = 0\]
On Solving for the determinant by \[{R_1}\], we get
⇒\[a\alpha ({r_2} - {r_1})(1 - {r_1})(1 \times (1 \times (1 + {r_1}) - 1 \times ({r_2} + {r_1}))) = 0\]
On Solving for the brackets in LHS, we get
⇒\[a\alpha ({r_2} - {r_1})(1 - {r_1})(1 + {r_1} - {r_2} - {r_1}) = 0\]
On Canceling out equal and opposite terms in LHS, we get
⇒\[a\alpha ({r_2} - {r_1})(1 - {r_1})(1 - {r_2}) = 0\]
Hence, if the solution needs to be 0, either \[({r_2} - {r_1}) = 0\] or \[(1 - {r_1}) = 0\] or \[(1 - {r_2}) = 0\]
Hence, either
⇒\[{r_2} = {r_1}\]
Or
⇒\[{r_1} = 1\]
Or
⇒\[{r_2} = 1\]

Note:
While solving determinant, we should always try to use the transformations in such a way that common terms can be taken out of it. This makes solving the determinants very easy and may also result in having two columns or rows the same, Hence resulting in 0.