Have You Tried Swapping The Positions?

Algebra Level 3

$\large{ \begin{cases} ax+a^2y + z = 0 \\ bx+b^2y + z = 0 \\ cx + c^2 y + z = 0 \end{cases}}$

If the system of equations above has exactly one trivial solution, $(0,0,0)$ , and that the constant non-zero coefficients $a,b,c$ follows a geometric progression then the common ratio cannot be $\text{\_\_\_\_\_\_}$ .

-3 -2 -1 2 3

1 solution

Venkata Karthik Bandaru
Mar 21, 2016

Solution :

But wait, what we found in the above picture was that the common ratio must be 1 or (-1) to ensure that the given system has more solutions than just the trivial one $\left( 0,0,0 \right)$ , and this is just the opposite of what we wanted. Hence, as per the problem statement, the values of the common ratio must not be 1 or (-1).

Moderator note:

Always be careful when dividing by an algebraic expression. You should state that either $b -c = 0$ or $ab+ac - a^2 - c = 0$ .

Similarly, we subsequently have $a = 0$ or $\frac{ \gamma + \gamma^2 - 1 - \gamma^2 } { \gamma} = 0$ .

Have You Tried Swapping The Positions?

1 solution

Moderator note:

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