I was reading random exercises, and found a typical
Determine what values of $a$ cause the system to have no solution, an unique solution and infinite solutions. Also find the solution set for each case. $$\begin {bmatrix} 3 & 0 & 3 & 1 \\ 0 & a & 2 & 2 \end {bmatrix}$$
I was looking at the answer, and it was pretty clear:
He first transformed it to
$$\frac{1}{3}r_1$$
$$\begin {bmatrix} 1 & 0 & 1 & 1/3 \\ 0 & a & 2 & 2 \end {bmatrix}$$
And then determined:
- When it has no solutions: Never - this system will always have a solution because regardless of the value of $a$, the system is consistent.
- When it has unique solution: Never - the rank of this system will always be lower than the number of columns, so this system can't have an unique solution.
And then, since its clear that the system must have infinite solutions, the guy begun calculating the solution set. I omit the procedure, but what he essentially did was:
- Have a case where $a = 0$. Replacing $a$ by a $0$ in the matrix, then reducing it to its echelon form, the guy found a set of infinite solutions.
- Have another case where $a \neq 0$. Reducing the matrix to its echelon form, the guy found ANOTHER set of infinite solutions.
As you can observe, the guy decided to make two cases. One for when $a$ is $0$ and another when it is not - achieving two different infinite solution sets.
What is not entirely clear to me is, what reasoning drove the man to choose these two cases? That is, what made him check for the number $0$ rather than, dunno, $2$ instead?