# gauss elimination with parameters

I'm breaking my head trying to figure out what i'm doing wrong,
i'm support to eliminate the following equations using Gauss elimination system

ax+y+z=4
x+by+z=3
x+2by+z=4
and i'm suppose to figure out for which a,b (which are Parameters) the system has One solution/Infinit Solutions/No solutions - if there's infint solutions i'm suppose to write the general solution for the system.
now it's all pretty basic in the begining

1    b  1  3
1  2b  1  4 R2->R2-R1
a    1  1  4 R3->R3-aR1

1   b           1  3
0   b           0  1       * i can reach the first value using /:b , i then get y=1/b
0  1-ab   1-a  4-3a

from here i'm puzzled as there's no way to carry on the elimination (if i'm doing it right)
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Commented:
You are making things too complicated.

What happens if   a = b = 1 ??

What values will make the 1st and 3rd equations wierd??
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Author Commented:
I dont see how it's possible, it will give the following result for the 3rs equation:
0 0 0 0 1 which is not possible, not as a solution anyway.
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Commented:
What happens to the 1st and 2nd equations if   a = b = 1 ??
You don't have to solve them, just look at them.

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Author Commented:
yes, it works out... but it doesnt for the 3rd so i cannot use those values as answer.
and i have to define if for those two parameters there's uniqe solution, unlimited solutions or no solutions (like if b=0 then 1/0 on the second equation isnt possible)
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Commented:
There are three cases to consider:

1st and 2nd    ==>   a = b = 1    ==>  No solutions

1st and 3rd    ==>   a=1  b=0.5  ==>  Infinite solutions

2nd and 3rd   ==>  b=0 (which you found)  ==>  No solutions

I'm pretty sure that all other cases have single solutions.
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Commented:
Whenever you have an equation involving a product like
r * s = t

When t is zero, at least one of r and s must be zero
When t is nonzero, both r and s must be nonzero

Also, whenever you want to eliminate a factor like in
a*b = 1
to get
a = 1/b
You simply cannot do that unless you have proven that b is nonzero.

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