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Math Help - Linear combinations - matrices

  1. #1
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    Linear combinations - matrices

    I'm confused about a problem for my matrix algebra class. The directions say to "Determine if b is a linear combination of a1, a2, and a3."

    The problem I am confused on is:

    (These are all 3 x 1 matrices, I don't know how to write out a matrix on the computer)

    [1]
    [-2] = a1
    [0]

    [0]
    [1] = a2
    [2]

    [5]
    [-6] = a3
    [8]

    [2]
    [-1] = b
    [6]

    The next thing I did was I used the equation (x1)(a1) + (x2)(a2) + (x3)(a3) = b, then came up with the system of equations:

    x1 + 0 + 5(x3) = 2
    -2(x1) + x2 - 6(x3) = -1
    0 + 2(x2) + 8(x3) = 6

    I put those in an augmented matrix:
    (3x4 matrix)

    [ 1 0 5 | 2]
    [-2 1 -6 | -1]
    [ 0 2 8 | 6]

    I used elementary row operations and got the matrix down to:

    [ 1 0 5 | 2 ]
    [ 0 1 4 | 3 ]
    [ 0 0 0 | 0 ]

    I don't know where to go from here. I don't think that b is a linear combination, but my professor only gave one example on how to do this and it was a little different than the one on the homework.
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  2. #2
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    Does the system, as you have reduced it, have at least one solution?
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    Does the system, as you have reduced it, have at least one solution?
    I don't think it does... I tried using Row Operations to turn the 5 and the 4 into zeroes, but I couldn't come up with a way to do that, so, no, I don't think it has a solution. If there is no solution, it means that b is NOT a linear combination of a1, a2, and a3, right?
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  4. #4
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    Well, you've done Gaussian elimination. Presumably, if you were going to find a solution, you'd do back substitution. What do you get when you try that?
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  5. #5
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    Quote Originally Posted by Ackbeet View Post
    Well, you've done Gaussian elimination. Presumably, if you were going to find a solution, you'd do back substitution. What do you get when you try that?
    Well, the system of equations I would have from the reduced matrix would be

    x1 + 5(x3) = 2
    x2 + 4(x3) = 3

    right? ... So,

    x1 = 2 - 5(x3)
    x2 = 3 - 4(x3)

    Unless I'm missing something, I don't think you can solve it by using substitution, since there are two equations and three variables.
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  6. #6
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    With any linear system, the number of solutions is either zero, one, or infinite. Because the system is underdetermined (more variables than equations), you can rule out the system having one solution. Do you think this system has zero solutions, or infinitely many solutions?
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  7. #7
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    Quote Originally Posted by Ackbeet View Post
    With any linear system, the number of solutions is either zero, one, or infinite. Because the system is underdetermined (more variables than equations), you can rule out the system having one solution. Do you think this system has zero solutions, or infinitely many solutions?
    Infinitely many because the last equation in the system would be, based on the reduced matrix,

    0(x1) + 0(x2) + 0(x3) = 0, and any number, when plugged in for x1, x2, and/or x3 would satisfy this equation.
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  8. #8
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    Your conclusion is correct, but not because of the correct reason. The system has infinitely many solutions because in your reduction down to

    x1 = 2 - 5(x3)
    x2 = 3 - 4(x3),

    you can have any value of x3, and x1 and x2 will be determined. The system would have no solutions if you had a system like this after row reduction:

    \left[\begin{matrix}1 &0 &3\\ 0 &1 &4\\ 0 & 0 &0\end{matrix}\left|\begin{matrix}3\\ 5\\ 2\end{matrix}\right]

    The last row there is a contradiction, because if you multiply every variable by zero, you're going to get zero, which is not equal to 2.

    In any case, the important thing in your case is that there is a solution. Getting back to the original question, what does this tell you?
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  9. #9
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    Quote Originally Posted by Ackbeet View Post
    Your conclusion is correct, but not because of the correct reason. The system has infinitely many solutions because in your reduction down to

    x1 = 2 - 5(x3)
    x2 = 3 - 4(x3),

    you can have any value of x3, and x1 and x2 will be determined. The system would have no solutions if you had a system like this after row reduction:

    \left[\begin{matrix}1 &0 &3\\ 0 &1 &4\\ 0 & 0 &0\end{matrix}\left|\begin{matrix}3\\ 5\\ 2\end{matrix}\right]

    The last row there is a contradiction, because if you multiply every variable by zero, you're going to get zero, which is not equal to 2.

    In any case, the important thing in your case is that there is a solution. Getting back to the original question, what does this tell you?
    Since there is a solution, then that means that b is a linear combination of the 3 matrices listed.

    How would I write that out, though? In the example my professor gave us, we ended up with something like x1 = 3, x2 = 4, which implied that 3(a1) + 4(a2) = b, and so b was a linear combination. But since this has infinite solutions, would I just write "infinite solutions - b is a linear combination" ?
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  10. #10
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    Well, I'd gather together all the info you have. You started with

    (x1)(a1) + (x2)(a2) + (x3)(a3) = b.

    You got it down to

    x1 = 2 - 5(x3)
    x2 = 3 - 4(x3).

    I would pick a value for x3. Find x1 and x2, and then simply rewrite

    (x1)(a1) + (x2)(a2) + (x3)(a3) = b.

    Make sense?
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  11. #11
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    Quote Originally Posted by Ackbeet View Post
    Well, I'd gather together all the info you have. You started with

    (x1)(a1) + (x2)(a2) + (x3)(a3) = b.

    You got it down to

    x1 = 2 - 5(x3)
    x2 = 3 - 4(x3).

    I would pick a value for x3. Find x1 and x2, and then simply rewrite

    (x1)(a1) + (x2)(a2) + (x3)(a3) = b.

    Make sense?
    Yes, that makes sense. Thank you so much for walking me through it! It helped a lot
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  12. #12
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    Good! You're welcome. Have a good one!
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