Question

If $a,b,c$ are in G.P with common ratio $r_1$ and $\alpha ,\beta ,\gamma$ are in G.P. with common ratio $r_2$, and the equations $ax+\alpha y+z=0,bx+\beta y+z=0,cx+\gamma y+z=0$ have only trivial solution, then which of the following is NOT true?

• A. $r_1\ne 1$
• B. $r_2\ne 1$
• C. $r_1\ne r_2$
• D. none of these

Answer 1

The correct option is D none of these

The given system of equations can be written as $AX=O$
where $A=\left[\begin{array}{ccc}a& \alpha & 1\\ b& \beta & 1\\ c& \gamma & 1\end{array}\right]$, $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],O=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$
Given a,b,c are in G.P with common ratio $r_1$
Then $b=a{r}_1,c=b{r}_1=a{r_1}^2$
Also, given $\alpha ,\beta ,\gamma$ are in G.P with common ratio $r_2$
Then $\beta =\alpha r_2,\gamma =\beta r_2=\alpha {r_2}^2$
Now for trivial solution,
$D\ne 0$
$\Rightarrow \left|\begin{array}{ccc}a& \alpha & 1\\ a{r}_1& \alpha r_2& 1\\ a{r}_1^2& \alpha r_2^2& 1\end{array}\right|\ne 0$
$\Rightarrow a\alpha \left|\begin{array}{ccc}1& 1& 1\\ r_1& r_2& 1\\ r_1^2& r_2^2& 1\end{array}\right|\ne 0$
$R_1\to R_1-R_2,R_2\to R_2-R_3$
$\Rightarrow a\alpha \left|\begin{array}{ccc}1-r_1& 1-r_2& 0\\ r_1(1-r_1)& r_2(1-r_2)& 0\\ r_1^2& r_2^2& 1\end{array}\right|\ne 0$
$\Rightarrow a\alpha (r_2-r_1)(1-r_1)(1-r_2)\ne 0$
$\Rightarrow r_2\ne r_1,r_1\ne 1,r_2\ne 1$
Hence, options A, B and C are true.
But we need to find the option which is not true. Hence, the correct answer is D.