Find (a,b) for a system of linear equations given

[LTP]

http://peecee.dk/uploads/042009/Screenshot4.png

I have to find (a,b) \in R when the system of linear equations
a) have no solution
b) have one solution
c) have infinitely mane solutions.

I've converted the equations so they're on a row echelon form. My question is, what do I do from here?

 

[HallsofIvy]

Do you mean that the equations are:

x₁+ (a-1)x₂+ 2x₃+ (a+2)x₄= a+ b
x₁+ 2ax₂+ ax₄= 2a+ b
-(a+1)x₂+ (2a+2)x₃= 0
(2a+2)x₂+ (4a-4)x₃+ (a^2+ a- 8)x₄= 4a+ ab+ b?

First, those are not in "row echelon form". There should not be an "x₁" in the second equation: subtract the first equation from the second equation to eliminate that x₁. Secondly, there should be no "x₂" or in the third equation nor should there be either an "x₂" nor an "x₃" in the fourth equation. Once you have used row operations to get rid of those your last equation will be "f(a,b)x₄= g(a,b)" where f and g are functions of a and b. There will be NO solution if g(a,b)= 0 while f(a,b) is not 0. There will be an infinite number of solutions if f(a,b)= 0 and g(a,b)= 0. Finally, there will be a unique solution if f(a,b) is not 0.

 

[LTP]

No, the picture just shows the system of equations.

This is it: http://peecee.dk/uploads/042009/Screenshot-1.png

Should I reduce it even further to the "reduced row echelon form", if I want to use your suggestion?

 

[LTP]

Ok, so I found that (a^2+a)x_4 = ab+b (ignore #3, I made a miscalculation).

a^2+a = 0 => a=0 or a=-1

No solution:
a^2+a=0 and ab+b != 0
a=0, b \in R except 0

One solution:
ab+b != 0 <=> ab != -b
a \in R except -1 and b \in R or the opposite.

No unique solution:
a^2+a=0 and ab+b = 0
a=0 and b= 0 or a=-1 and b \in R

Is this correct?