Linear Algebra (Conditions for solutions).

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SUMMARY

The discussion focuses on the conditions for solutions of the linear system defined by the equations x + by = -1 and ax + 2y = 5. It establishes that if ab ≠ 2, there is a unique solution given by x = (5b + 2)/(ab - 2) and y = (5 + a)/(2 - ab). If ab = 2 and a ≠ -5, there are no solutions due to the inconsistency of the system. Conversely, if ab = 2 and a = -5, the system has infinitely many solutions, represented by the parametric equations x = 2t/5 - 1 and y = t.

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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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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[/color], 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.
 
Brilliant. Thanks a million for that!
 

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