Matrix (more or less theoretical)

[Melawrghk]

Homework Statement

If a linear system has the augmented matrix:
a, -b, 4, 0
1, -2, 1, 0
0, 0, 4, 0

then select all correct answers:
(a)The system has infinitely many solutions if b=2a;
(b)The system is inconsistent if b=2a;
(c)The system is consistent if b=a;
(d)The system has exactly one solution if a=b=0;
(e)The system has exactly one solution if 2a doesn't equal "b"

Homework Equations

None

The Attempt at a Solution

So my prof didn't explain this topic at all, but it's in an assignment that is due on Sunday night...
I've tried my own approach, but somehow I don't get the right answers...
I did matrices separately for the 3 cases: "b=2a", "b=a" and "a=b=0"
Case 1: b=2a
I won`t write out the whole process (because its hideous) but here is the matrix I got in the end:
1, 0, 0, 0
0, 1, -2, 0
0, 0, 4a, 0
So I figured this would mean that the system would have an infinite number of solutions. Thus, if b didn`t equal 2a, then the system should have a unique solution. So I eliminated (b) from the list.
Case 2: b=a
This time I got:
1, 0, 0, 0
0, 1, -2, 0
0, 0, -a, 0
And that would mean that the system has an infinite number of solutions, once again. So (c) holds up as true.
Case 3: a=b=0
1, 0, 0, 0
0, 1, -2, 0
0, 0, 0, 0
Which would once again mean that it has an infinite number of solutions. This would make (d) false

So is my logic right at all? I honestly don`t know what else to do.. Thanks in advance!

 

[HallsofIvy]

I don't know why you say the process is "hideous". After just two steps (swap first and second rows; subtract a times first row from second) I get
[tex]\left[\begin{array}{cccc}1 & -2 & 1 & 0 \\ 0 & 2b-a & 4-a & 0 \\ 0 & 0 & 4 & 0\end{array}\right][/tex]
which tells you pretty much what you want to know.

Without even doing that, since the last column is all zeros, you know immediately that the system is NEVER inconsistent. As long as the "pivots", the numbers on the main diagonal, are not 0, the system has exactly one solution. If any of the pivots are 0, the system has infinitely many solutions.

 

[Melawrghk]

Okay, thanks :) I thought it was hideous because I don't know how to make matrices look pretty on computer screens...

But what about this:
7, 1, 1, 3
1, 7, 7, a
1, 7, 1, b

(a)the system has a unique solution for some but not all a and b
(b)the system is inconsistent for all a and b
(c)the system has a unique solution for all a and b
(d)the system is consistent for all a and b
(e)the system is inconsistent for a=b

So I reduced it to:
1, 7, 7, a
0, 1, 1, (3-7a)/48
0, 0, 1, -(b-a)/6

So the lack of zeros in the pivots would mean it's got a single solution, right? That would eliminate (b) and (e). (a) kind of suggests that it might not have a unique solution, so that would eliminate (a) as well. That would leave (c) and (d). Is that right? Or did I just miss the point again?

Thanks tons for your help!

 

[Melawrghk]

Actually, never mind about that one, I got it.

But, maybe you could help me with this one?
x+y=z
2y+z=x
z+2x=y
The system can be written as AX=B where B has constant entries. Select all correct answers from below:
(a)A is a 3x3 matrix and b=[0 0 0]^T
(b)A is a 3x2 matrix and b=[z x y]^T
(c)A is a 3x2 matrix and b=[x y z]^T
(d)A is a 3x3 matrix and b=[z x y]^T
(e) none of the above

I was able to eliminate (c) because its b^T doesn't equal b in the example. Then, I think A is a 3x2 matrix, so I got rid of (a) and (d). So now, for (b) I'm thinking maybe it's not true because the question says B has constant entries, meaning no variable and things of that nature. So that would mean the only right answer is (e)...

Am I right?

 

[HallsofIvy]

Yes. If the determinant of a matrix is not 0 (which is the same as saying that none of the pivots of the reduced matrix are 0), then any equation with that matrix as coefficent matrix has a unique solution. The only time either "infinitely many solutions" or "no solutions" can happen is if you get a 0 on the diagonal. And then the right hand side becomes important.

 

[Melawrghk]

Solved :)