Liner algebra- existence and uniqueness

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Homework Statement


Coeffcient Data and Existence and Uniqueness of Solutions. Assuming that a (not equal to) 0, and an equation that restricts a; b; c; d so
that the following system has only the trivial solution.
(1) ax1 + bx2 = 0
(2) cx1 + dx2 = 0

Hint: Find the echelon form of the associated matrix equation and from this echelon form and a relation/rule involving a; b; c; d so that the
augmented matrix has as many pivots as variables. If you have taken the determinant of 2 x 2 matrices then you can check your work.

Homework Equations





The Attempt at a Solution



I have no idea how to do this problem, I'm not completely sure what its asking of me. If someone could guide me through it that would really help.
 

Answers and Replies

  • #2
LCKurtz
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Reduce the matrix

[tex]\left(\begin{array}{cc}
a & b\\
c & d
\end{array}\right)
[/tex]

to the identity by row operations. Remember you can't divide by zero so that may give you some conditions. You are already given that a isn't zero.
 
  • #3
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[a b | 0;
c d | 0]

[a b | 0;
c-(c/a)c d-(d/b)d | 0]

that doesn't seem right to me, i'm not sure what else I can do with it though....

am i at least on the right track?
 
  • #4
vela
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I'm not sure what you did there, but it doesn't look like an elementary row operation.

To reduce the matrix, the first thing you want to do is turn the a into a 1, so you'd divide the first row by a, which you can do since you're given [itex]a\ne 0[/itex]. This would leave you with

[tex]\begin{pmatrix}1 & b/a & 0 \\ c & d & 0\end{pmatrix}[/tex]

Do you see where to go from here?
 
  • #5
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oooh, sorry I didn't even think of that. Would i row reduce from here? Try to get b to equal zero?
 
  • #6
vela
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oooh, sorry I didn't even think of that. Would i row reduce from here?
Yes, you would continue to row reduce. That's what the hint is telling you to do when it says to find the echelon form of the matrix. What should be your next step in getting to echelon form?

Try to get b to equal zero?
No, you should consider b a given value. You can't do anything to make b=0.
 
  • #7
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Yes, you would continue to row reduce. That's what the hint is telling you to do when it says to find the echelon form of the matrix. What should be your next step in getting to echelon form?


No, you should consider b a given value. You can't do anything to make b=0.
oops sorry, i meant to say c. So i could do r2 = r2 - cr1 and get 0 where c is and d-(b/a)*c where d is
 
  • #8
vela
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oops sorry, i meant to say c.
Oh, okay, what you meant by "b" was the upper-right entry in the matrix, not b itself. Likewise, you're not setting c to 0 now but the lower-left entry.
So i could do r2 = r2 - cr1 and get 0 where c is and d-(b/a)*c where d is
Right, so now you have the matrix

[tex]
\begin{pmatrix}1 & b/a & 0 \\ 0 & d-(b/a)c & 0\end{pmatrix}
[/tex]

which is now in echelon form. What does the rest of the hint tell you?
 
  • #9
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well i'm trying to find how to restrict it to where the trivial solution is the only solution but i'm not sure how to do this. It already has as many pivots as variables as long as d-c(b/a) isn't zero. I don't know how I would restrict it to only have a trivial solution.
 
  • #10
vela
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As you noted, if d-c(b/a)≠0, you'll have a second pivot, and the reduced matrix will be

[tex]
\begin{pmatrix}1 & b/a & 0 \\ 0 & 1 & 0\end{pmatrix}
[/tex]

Once you have a reduced matrix, you can solve the system. What's the solution to this system? What turns out to be different if d-c(b/a)=0 so you don't get that second pivot?
 
  • #11
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As you noted, if d-c(b/a)≠0, you'll have a second pivot, and the reduced matrix will be

[tex]
\begin{pmatrix}1 & b/a & 0 \\ 0 & 1 & 0\end{pmatrix}
[/tex]

Once you have a reduced matrix, you can solve the system. What's the solution to this system? What turns out to be different if d-c(b/a)=0 so you don't get that second pivot?
so... x+ (b/a)y = 0
and y = 0 is the solution to that matrix. whats different is if you don't have a 2nd pivot you have a free variable?
 
  • #12
vela
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so... x+ (b/a)y = 0 and y = 0 is the solution to that matrix.
Well, those are the equations corresponding to matrix, not the solution. The solution is x=y=0.
whats different is if you don't have a 2nd pivot you have a free variable?
Yes, which means what about the possible solutions to the system?
 
  • #13
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Well, those are the equations corresponding to matrix, not the solution. The solution is x=y=0.

Yes, which means what about the possible solutions to the system?
when it has a free variable it can have another solution other than the trivial
 
  • #14
vela
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Exactly.
 
  • #15
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Thank you so much for all the help. I actually understand how ans why this works. Thanks again
 

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