# Matrix with 2 unknowns

1. Sep 19, 2013

### Temp0

1. The problem statement, all variables and given/known data
In the following problem, find conditions on a and b such that the system has no solution, one solution, and infinitely many solutions.

x + by = -1
ax + 2y = 5

2. Relevant equations
None that I know of.

3. The attempt at a solution
Basically, first I put the entire equation into a matrix.
[ 1 b | -1
a 2 | 5 ]
I reduce the bottom by subtracting R2 - aR1
[ 1 b | -1
0 2-ab| 5 +a ]
I then reduce the bottom again by dividing R2/(2-ab)
[ 1 b | -1
0 1 | (5+a)/(2-ab)]
I remove the b from the top by subtraction: R1 - bR2
[1 0 | -1 - b((5+a)/(2-ab))
0 1 | (5+a)/(2-ab) ]
This leaves me with the values for x and y, and for the first question I am correct in saying if ab = 2, then there is no solution as it is undefined. However, my unique solution is somehow wrong and I would like some help in determining if I made an error or I somehow didn't reduce something.

The correct unique solution is: x = (-2 - 5b)/(2-ab) y = (a+5)/(2-ab)

Also, I have no idea what finding an infinite solution means, I would really like some help on clarifying that. Thank you.

2. Sep 19, 2013

### tiny-tim

Hi Temp0! Welcome to PF!
That is the same as yours!
Hint: if A and B are two solutions, what can you say about A - B ?

3. Sep 19, 2013

### lendav_rott

For example: if you are given graphs of f(x) and g(x) [neither contain discontinuities] and g(x) 's graph never "falls under" or crosses f(x) , no matter what the argument, g(x) has always higher values than f(x) and you are asked to provide solutions for g(x) > f(x) then you can say that there are infinitely many solutions. But you are dealing with an equal sign so that must mean the 2 graphs are...?

4. Sep 20, 2013

### Temp0

Hmm, thanks for the help on the infinite thing, I finally get that ^^. However, i've tried reducing and expanding my answer, but it never becomes the same as the answer in the book. Are there any hints you can give me? =D

edit: nvm, I just looked at it again and realized how to get to the answer, thanks for your help =p.