Finding possible coefficients to a system of linear equations

Question

I have this system of linear equations

$$\begin{cases} x_1+x_2+3x_3=2 \\ x_1 + 2x_2 + 4x_3 = 3 \\ x_1 + 3x_2 + \alpha x_3 = \beta \\ \end{cases}$$

And I have to find the values of $\alpha$ and $\beta$ that makes the system:

1. Have a single solution
2. Have multiple solutions
3. Have no solutions

I'm trying to use the matrix of the coefficients to calcule the determinant and I'm ending with $\alpha - 5$, meaning that $\alpha$ should be not 5 to make the system have a single solution, but messing around with a calculator online I found that actually, $\alpha$ needs to be different from 6 to make the system have a single solution, regardless the $\beta$ value.

To make the system have multiple solutions, I also found that $\alpha = 6$ and $\beta = 5$, and to make it have no solutions, $\alpha = 6$ and $\beta \neq 5$.

My question is how to find those values? Because apparently the method I am trying does not work.

Answer 1

A single solution is guaranteed when the LHS are linearly independent.

By subtracting the first equation from twice the second, we get

$$x_1+3x_3+5x_3=4,$$ the LHS of which coincides with the third when $\alpha=5$.

Then, if $\beta=4$, the RHS also coincide and the system is compatible.

Answer 2

Hint: Your system equation can be writen by $\begin{cases} x_1+ x_2 + 3x_3& =2\\ \quad \quad x_2 + x_3& =1\\ \quad (\alpha - 5) x_3 &= (\beta -4) \end{cases}$.