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.


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.

  • $\begingroup$ So, the values that I got from the online calculator are wrong and the way that I was originally going was right? $\endgroup$ – Luiz Eduardo Simões Dec 20 '16 at 23:26
  • $\begingroup$ Just confirming that the values from the online calculator were wrong, I used another website to test and I got exactly those values. $\endgroup$ – Luiz Eduardo Simões Dec 22 '16 at 12:20

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


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