MHB Finding unknown coefficients to give a unique solution in a linear system

[qybah]

Hi can someone please explain how to do this question:
Given two equations;

x-ay= 1

ax-4y=b

For which values of a does each system have a unique solution, and for which pairs of values (a,b) does each system have more than one solution?

All help is greatly appreciated

 

[Fernando Revilla]

Using the elemental transformation $E_2\to E_2-aE_1:$
$$\left \{ \begin{matrix} \begin{aligned} &x-ay=1 \\& ax-4y=b\end{aligned}\end{matrix}\right.\sim \left \{ \begin{matrix} \begin{aligned} &x-ay=1 \\& \;\;(a^2-4)y=b-a\end{aligned}\end{matrix}\right.$$ Now, analize the cases $a^2-4=0$ and $a^2-4\ne 0.$

 

[qybah]

Using the elemental transformation $E_2\to E_2-aE_1:$
$$\left \{ \begin{matrix} \begin{aligned} &x-ay=1 \\& ax-4y=b\end{aligned}\end{matrix}\right.\sim \left \{ \begin{matrix} \begin{aligned} &x-ay=1 \\& \;\;(a^2-4)y=b-a\end{aligned}\end{matrix}\right.$$ Now, analize the cases $a^2-4=0$ and $a^2-4\ne 0.$

Hi thanks for the reply. Am i on the right track if i do it like this?

Also if i were to analyze the two cases you mentioned, would it be correct to say that when $a^2-4\ne 0$ or $a^2\ne\pm 2$ the system has a unique solution? Thanks in advance

 

[Fernando Revilla]

Hi thanks for the reply. Am i on the right track if i do it like this?

Right, but it remains to be seen several cases. You should obtain

$a\ne \pm 2,$ unique solution.

$a=2\left \{ \begin{matrix} b=2 &\mbox{more than one solution }& \\b\ne2 & \mbox{no solution}\end{matrix}\right.$

$a=-2\left \{ \begin{matrix} b=-2 &\text{more than one solution}& \\b\ne -2 & \mbox{no solution}\end{matrix}\right.$

Also if i were to analyze the two cases you mentioned, would it be correct to say that when $a^2-4\ne 0$ or $a^2\ne\pm 2$ the system has a unique solution? Thanks in advance

I suppose you meant $a\ne \pm 2.$ In that case, you are right.