Find (a,b) for a system of linear equations given

  • Thread starter LTP
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  • #1
LTP
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http://peecee.dk/uploads/042009/Screenshot4.png

I have to find (a,b) \in R when the system of linear equations
a) have no solution
b) have one solution
c) have infinitely mane solutions.

I've converted the equations so they're on a row echelon form. My question is, what do I do from here?
 

Answers and Replies

  • #2
HallsofIvy
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Do you mean that the equations are:

x1+ (a-1)x2+ 2x3+ (a+2)x4= a+ b
x1+ 2ax2+ ax4= 2a+ b
-(a+1)x2+ (2a+2)x3= 0
(2a+2)x2+ (4a-4)x3+ (a^2+ a- 8)x4= 4a+ ab+ b?

First, those are not in "row echelon form". There should not be an "x1" in the second equation: subtract the first equation from the second equation to eliminate that x1. Secondly, there should be no "x2" or in the third equation nor should there be either an "x2" nor an "x3" in the fourth equation. Once you have used row operations to get rid of those your last equation will be "f(a,b)x4= g(a,b)" where f and g are functions of a and b. There will be NO solution if g(a,b)= 0 while f(a,b) is not 0. There will be an infinite number of solutions if f(a,b)= 0 and g(a,b)= 0. Finally, there will be a unique solution if f(a,b) is not 0.
 
  • #3
LTP
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No, the picture just shows the system of equations.

This is it: http://peecee.dk/uploads/042009/Screenshot-1.png

Should I reduce it even further to the "reduced row echelon form", if I want to use your suggestion?
 
  • #4
LTP
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Ok, so I found that (a^2+a)x_4 = ab+b (ignore #3, I made a miscalculation).

a^2+a = 0 => a=0 or a=-1

No solution:
a^2+a=0 and ab+b != 0
a=0, b \in R except 0

One solution:
ab+b != 0 <=> ab != -b
a \in R except -1 and b \in R or the opposite.

No unique solution:
a^2+a=0 and ab+b = 0
a=0 and b= 0 or a=-1 and b \in R

Is this correct?
 

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