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Linear Algebra - Linear Systems and Matrices

  1. Jan 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Suppose we know that a linear system Ax = b has a unique solution. What can we say about the solutions of the linear system Ax = 0?

    a) It has the same solution.
    b) The solution to Ax = b is also a solution to Ax=0, but there may be other solutions.
    c) Ax = 0 has a unique solution, but it may be different from the solution to Ax = b.
    d) The solution to Ax = b is also a solution to Ax = 0, but the latter has one more solution, namely x= 0.
    e) Ax=0 has infinitely many solutions, but the solution to Ax=b need not be one of them.
    f) none of the above.

    2. Relevant equations



    3. The attempt at a solution

    I kinda understand the problem, but not sure what the multiple choice options meant.
    For now I think the correct option is option B. I don't understand what is the difference between this option (b) and option (d), what is x=0? Does it mean a system of 0x+0y+... and etc? If so then this is false since it has more than just this it could have ifinitely many solutions not just 0x+0y...? I am confusing myself I think...
     
  2. jcsd
  3. Jan 27, 2013 #2

    Zondrina

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    So you know there's a unique solution to Ax = b. Call this unique solution p so that Ap = b.

    Now since you know p is a unique solution, what can you say about the linear system itself in terms of its dependence?
     
  4. Jan 27, 2013 #3
    what do you mean by dependence?
     
  5. Jan 27, 2013 #4

    Zondrina

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    Is your system linearly independent or linearly dependent?
     
  6. Jan 27, 2013 #5
    what is dependent or independent??? It didnt say what the system was...
     
  7. Jan 27, 2013 #6

    Zondrina

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    Ahh I see, so you're not familiar with those concepts yet. I'll try to explain them as lightly as possible.

    So linear independence means when you solve your system, there is going to be a leading 1 in every column ( I'm sure you HAVE to be familiar with leading 1 ) yielding a unique solution to your system.

    Linear dependence means when you solve your system, there wont be a leading 1 in every column which will actually result in infinitely many solutions.

    Could you tell me anything now?
     
  8. Jan 27, 2013 #7
    the question says nothing about that only says that Ax=b has an unique solution.
     
  9. Jan 27, 2013 #8

    Zondrina

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    Perhaps you should read more close :

    I'll even go as far as to point out if you still haven't seen it. Since your system has that unique solution 'p' I was talking about earlier, it means your system is LINEARLY INDEPENDENT.

    That means when you row reduce your co-efficient matrix A, there will be a leading 1 in every column no matter what.

    Now what happens if we replace b with the zero vector and solve Ax = 0?

    There's going to be a leading 1 in every column no matter what right? What happens when you row reduce the zero vector though?
     
  10. Jan 27, 2013 #9
    There is going to be infinite number of solutions? They all = 0? So I am assuming e is correct?
     
    Last edited: Jan 27, 2013
  11. Jan 27, 2013 #10

    Zondrina

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    This statement makes me think you're not visualizing at all. Perhaps I should give you an example. Consider this co-efficient matrix, lets call it A :

    A = [itex] \left( \begin{array}{cc}
    1 & 2 \\
    1 & 3 \\ \end{array} \right)[/itex]

    Lets take a vector of constants, call it b :

    b = [itex] \left( \begin{array}{c}
    1 \\
    2 \\ \end{array} \right)[/itex]

    Now I assure you, and you should verify it yourself, that the system Ax = b has a unique solution that is the ONLY solution to that system. Suppose that solution is 'p'.

    Now, suppose we replace b with the 0 vector : [itex] \left( \begin{array}{c}
    0 \\
    0 \\ \end{array} \right)[/itex]

    Solve the system Ax = 0 and compare your solutions. Could you tell me what's going on now?
     
  12. Jan 27, 2013 #11
    one is -1 and 1 the other is both 0s. So Ax=b is not a solution to ax=0.
     
  13. Jan 27, 2013 #12

    Zondrina

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    Not quite. I understand what you're trying to say, but be more clear.

    So you found the solution to Ax = b to be the vector [-1, 1].

    You also found the solution to Ax = 0 to be the vector [0, 0].

    Notice how for both of the systems, you were able to find a unique solution that was the ONLY solution.

    This occurred because the matrix A was linearly independent ( Has a leading 1 in every column when you row reduce ).

    Finally, notice that both your solutions aren't necessarily equal to each other?

    Now, look at the answers to the problem again and tell me. What do you know.
     
  14. Jan 27, 2013 #13
    I think c fits, since it says Ax=0 has a unique solution, but it may be different from the solution to ax=b. What you said there I know, but I think my english is horrible and did not understand the problem. Because when it said it may be different from the solution to Ax=b, i thought it also meant it should also be equal. Ok that sentence might be confusing...my english is bad...
     
  15. Jan 27, 2013 #14

    Zondrina

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    Perfectly fine, and also yes c is indeed the correct answer!

    If Ax = b has a unique solution call it 'p', we know that it is the only unique solution to Ax = b.

    This also informs us that the matrix A is LINEARLY INDEPENDENT, so that when we solve the system Ax = 0, we will also get a unique solution call it 'q'.

    As you have seen, it is not the case that the two solutions are equal, so clearly c is correct.

    As a side note, if your matrix A is linearly independent, and you are solving Ax = 0, then the only solution which is also unique is the zero vector.

    You'll probably learn more about that later though. Good job for now though.
     
  16. Jan 27, 2013 #15
    sorry I was playing tetris while doing this so my mind wasn't ... set for homework so I might of acted stupid. Next time work is work. Thanks for help
     
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