Given the system of equations:
$ x_1 + βx_2 + βx_3 + βx_4 = α $
$x_1 + (1 + β) x_2 + (α + β) x_3 + 2βx_4 = α + β$
$2x_1 + 2βx_2 + (α + 2β) x_3 + (α + 2β) x_4 = 3α$
$3x_1 + 3βx_2 + 3βx_3 + (α + 2β) x_4 = 4α + 2$
I need to find the values of $\alpha,\:\beta\:\in\mathbb{R}$ such as that the system has:
- a unique solution
- infinite solutions
- no solutions
Now obviously the first thing that comes to mind is using Gaussian elimination until a row echelon form is reached, but that seems exceptionally cumbersome with the parameters α and β. Is there a neater way to solve this that I'm missing?
By Gaussian-elimination,
$\begin{cases}x_1 + \beta x_2 + βx_3 + βx_4 = α \\ x_1 + (1 + β) x_2 + (α + β) x_3 + 2βx_4 = α + β\color{red}{(II-I)}\\2x_1 + 2βx_2 + (α + 2β) x_3 + (α + 2β) x_4 = 3α\color{red}{(III-2\times I)}\\3x_1 + 3βx_2 + 3βx_3 + (α + 2β) x_4 = 4α + 2\color{red}{(IV-3\times I)}\end{cases}$ $$\iff$$
$\begin{cases}x_1 + \beta x_2 + βx_3 + βx_4 = α \\ x_2 + α x_3 + βx_4 = β\\ α x_3 + α x_4 = 3α\\(α -β) x_4 = α + 2\end{cases}$
First case: $\alpha=0$. Then this system is equivalent to $$\begin{cases}x_1 + \beta x_2 + βx_3 + βx_4 = 0 \\ x_2 + βx_4 = β\\-β x_4 = 2\end{cases}$$ If $\beta=0$ then the system has no solution.
If $\beta\neq0$ then $$\begin{cases}x_1=2+2\beta-\beta^2-\beta x_3\\ x_2 = -2+β\\x_4 = -\frac{2}{\beta} \end{cases}$$
This system has infinite solutions.
Second case: $\alpha \neq 0.$
If $\beta=\alpha\neq-2$, there is no solution.
If $\beta=\alpha=-2$ then $\begin{cases}x_1-2 x_2 -2x_3 -2x_4 = -2\\ x_2 -2 x_3 -2x_4 = -2\\ x_3 + x_4 = 3\end{cases}$
This system has infinite solutions.
If $\beta\neq \alpha$, the system has a unique solution.