How to solve this parametric linear equation?

Question

How to solve this parametric linear equation? I need to find all numbers for $\alpha$ with which has a single, infinity or none solution.

$$ \left[\begin{array}{rrr|r} \alpha & 1 & 0 & 1 \\ 2\alpha & \alpha & 1 & 4 \\ 0 & 2 & 1 & 2 \end{array}\right] $$

I end up with this and don't know how to remove 2 from bottom row to have zeros bellow the diagonal.

$$ \left[\begin{array}{rrr|r} \alpha & 1 & 0 & 1 \\ 0 & \alpha-2 & 1 & 2 \\ 0 & 2 & 1 & 2 \end{array}\right] $$

Thank you.

Answer 1

Before anything, consider the case of $\alpha = 0$. Note that in this case, we can't get a pivot in the first column, and the system is $$ \left[\begin{array}{rrr|r} 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 4 \\ 0 & 2 & 1 & 2 \end{array}\right] $$ Reduce further to show that this will yield no solutions.

Now, following your steps, you've reached $$ \left[\begin{array}{rrr|r} \alpha & 1 & 0 & 1 \\ 0 & \alpha-2 & 1 & 2 \\ 0 & 2 & 1 & 2 \end{array}\right] $$ proceed by breaking the problem into two cases. The first (and easier) case is when $\alpha - 2 = 0$ (which is to say $\alpha = 2$). We have $$ \left[\begin{array}{rrr|r} 2 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \\ 0 & 2 & 1 & 2 \end{array}\right] \to \\ \left[\begin{array}{rrr|r} 2 & 1 & 0 & 1 \\ 0 & 2 & 1 & 2 \\ 0 & 0 & 1 & 2 \\ \end{array}\right] $$ which you can see leads to a unique solution.

Otherwise, we have $$ \left[\begin{array}{rrr|r} \alpha & 1 & 0 & 1 \\ 0 & \alpha-2 & 1 & 2 \\ 0 & 2 & 1 & 2 \end{array}\right] \to \\ \left[\begin{array}{rrr|r} \alpha & 1 & 0 & 1 \\ 0 & 1 & 1/(\alpha-2) & 2/(\alpha-2) \\ 0 & 2 & 1 & 2 \end{array}\right] $$ If $\alpha \neq 4$, we will have a unique solution.

Otherwise, the system reads as $$ \left[\begin{array}{rrr|r} \alpha & 1 & 0 & 1 \\ 0 & 1 & 1/2 & 1 \\ 0 & 2 & 1 & 2 \end{array}\right] $$ verify that this system yields infinitely many solutions.

Answer 2

The elementary row operations you performed are valid and don't change the solution set. You can do another one, though: swap rows 2 and 3, so you get $$ \left[\begin{array}{ccc|c} \alpha & 1 & 0 & 1 \\ 0 & 2 & 1 & 2 \\ 0 & \alpha-2 & 1 & 2 \\ \end{array}\right] $$ Now sum to the third row the second row multiplied by $-(\alpha-2)/2$, getting $$ \left[\begin{array}{ccc|c} \alpha & 1 & 0 & 1 \\ 0 & 2 & 1 & 2 \\ 0 & 0 & \frac{4-\alpha}{2} & 4-\alpha \\ \end{array}\right] $$ Now you can clearly see that, for $\alpha\ne0$ and $\alpha\ne4$, the system has unique solution (rank $3$ of the incomplete matrix). For $\alpha=4$, the system has infinitely many solutions. It remains to see what happens for $\alpha=0$: $$ \left[\begin{array}{ccc|c} 0 & 1 & 0 & 1 \\ 0 & 2 & 1 & 2 \\ 0 & 0 & 2 & 4 \\ \end{array}\right]\to \left[\begin{array}{ccc|c} 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 2 & 4 \\ \end{array}\right]\to \left[\begin{array}{ccc|c} 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 4 \\ \end{array}\right] $$ So, no solution.