Linear Algebra (Conditions for solutions).

In summary, the given system of equations has no solution if ab ≠ 2. If ab = 2 and a ≠ -5, there are no solutions. If ab = 2 and a = -5, the system has infinitely many solutions represented by the parametric equations x = 2t/5 -1 and y = t. The relevance of a ≠ -5 when ab = 2 can be seen in the reduced augmented matrix, where the bottom row is either all 0's or has a nonzero element, indicating the type of solution the system has.
  • #1
SherlockOhms
310
0

Homework Statement


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.


Homework Equations


General Algebra really.


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.
 
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  • #2
DAPOS said:

Homework Statement


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.

Homework Equations


General Algebra really.

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.
I used an augmented matrix.
DAPOS said:
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.
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
DAPOS said:
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.
 
  • #3
Brilliant. Thanks a million for that!
 

FAQ: Linear Algebra (Conditions for solutions).

1. What are the necessary conditions for a system of linear equations to have a unique solution?

The system must have the same number of equations as unknown variables, and the equations must be independent (i.e. none can be a multiple of another).

2. Can a system of linear equations have no solution?

Yes, if the equations are contradictory (i.e. one equation contradicts another), the system will have no solution.

3. How do we determine if a system of linear equations has infinitely many solutions?

If the number of equations is less than the number of unknown variables, there will be infinitely many solutions. This can also occur if one equation is a multiple of another.

4. Is it possible for a system of linear equations to have more than one unique solution?

No, a system of linear equations can either have a unique solution, no solution, or infinitely many solutions. It cannot have more than one unique solution.

5. What do we call the values that make a system of linear equations true?

The values that satisfy all the equations in a system are called solutions or roots of the system.

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