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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?

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  • $\begingroup$ @Rustyn: Yeah, well, the matrix I present is the augmented form already. $\endgroup$
    – Saturn
    Apr 8, 2014 at 3:45
  • $\begingroup$ OH thanks. sorry $\endgroup$
    – Rustyn
    Apr 8, 2014 at 3:45
  • $\begingroup$ @Amzoti: Well, the matrix had a different, original form. He then transformed it with row operations to the one I posted, and then drew all the conclusions based on that one, so I omitted the original form. I have added it now - but ultimately the number of rows is the same. $\endgroup$
    – Saturn
    Apr 8, 2014 at 4:38
  • $\begingroup$ @Amzoti: I have edited the post now (the original is the one in the quote box). $\endgroup$
    – Saturn
    Apr 8, 2014 at 4:46

1 Answer 1

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Case 1: Divide the first row by $3$ and the second row by $a$, yielding:

$$\begin {bmatrix} 1 & 0 & 1 & 1/3 \\ 0 & 1 & \dfrac{2}{a} & \dfrac{2}{a} \end {bmatrix}$$

Clearly, $a$ can be anything not equal to zero, so we have an infinite number of solutions.

Case 2: Set $a=0$, pre reduction, so we have the RREF:

$$\begin {bmatrix} 1 & 0 & 0 & -\dfrac{2}{3} \\ 0 & 0 & 1 & 1 \end {bmatrix}$$

A different set of infinite solutions.

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  • $\begingroup$ Yeah, that's the solution he wrote, but how did he know that he must test for $a = 0$ and $a \neq 0$ before actually doing the tests? Intuition? $\endgroup$
    – Saturn
    Apr 8, 2014 at 5:02
  • $\begingroup$ Another way to have the leading $1$ is by having $a = 1$. So why not check that instead? $\endgroup$
    – Saturn
    Apr 8, 2014 at 5:08
  • $\begingroup$ Yeah, I see it now, thanks. $\endgroup$
    – Saturn
    Apr 8, 2014 at 5:15
  • $\begingroup$ You are quite welcome. Regards $\endgroup$
    – Amzoti
    Apr 8, 2014 at 5:15
  • $\begingroup$ Five answers away from $1$K!!! $\endgroup$
    – amWhy
    Apr 8, 2014 at 11:54

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