# Homework Help: Linear Algebra (Conditions for solutions).

1. Apr 17, 2013

### SherlockOhms

1. The problem statement, all variables and given/known data
Consider the following system:
x + by = -1
ax + 2y = 5
Find the conditions on a and b such that the system has no solution, one solution or infinitely many solutions.

2. Relevant equations
General Algebra really.

3. The attempt at a solution
Previously we had been using augmented matrices to solve these sort of problems but using one here doesn't really help.
Rearranging the equations you get x = (5b + 2)/(ab - 2) and y =(5 + a) / (2 - ab)

The answer is:
If ab≠2 then the unique solution is x = (5b + 2)/(ab - 2) and y =(5 + a) / (2 - ab).
If ab = 2 and ab ≠ -5 then there are no solutions.
If ab =2 and a = -5 then x = 2t/5 -1 and y = t.

I understand why ab cannot be 2 for the first part of the answer but why is the fact that a can or cannot equal 5 when ab = 2 relevant for the second and third parts of the answer? Thanks.

2. Apr 17, 2013

### Staff: Mentor

I used an augmented matrix.
The above should be - If ab = 2 and a ≠ -5, then there are no solutions.

The reason for this can be seen in a reduced augmented matrix.
$$\begin{bmatrix} 1 & b & | & -1 \\ 0 & 2 - ab & | & 5 + a\end{bmatrix}$$

If ab = 2, the bottom row is 0 0 | 5 + a.

If a = -5, the bottom row is 0 0 | 0, which implies that there are an infinite number of solutions. Geometrically, the system represents two lines that coincide.

If a ≠ -5, the bottom row is 0 0 <nonzero>, which makes the system inconsistent. Geometrically, the system represents two parallel lines that do not coincide, hence don't intersect.
If a

3. Apr 17, 2013

### SherlockOhms

Brilliant. Thanks a million for that!

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