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Math Help - [SOLVED] Linear Equation

  1. #1
    Member ronaldo_07's Avatar
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    [SOLVED] Linear Equation

    Solve the system of linear equations in complex numbers:

    (2+i) x + (-3i) y = 3+i

    3i x + (3-2i) y = 3-3i

    How would I do this?
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  2. #2
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    Quote Originally Posted by ronaldo_07 View Post
    Solve the system of linear equations in complex numbers:

    (2+i) x + (-3i) y = 3+i

    3i x + (3-2i) y = 3-3i

    How would I do this?
    Equate the real and imaginary parts on each side:

    (2+i) x + (-3i) y = 3+i: Solve 2x = 3 and x - 3y = 1.
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  3. #3
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    you can easily solve linear equations like this 1 with Gaussian elimination
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  4. #4
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    Quote Originally Posted by 1234567 View Post
    you can easily solve linear equations like this 1 with Gaussian elimination
    If it's so easy why don't you solve it yourself? Mr.F provided a very simple approach to the problem and you should be able to easily solve from what he has provided.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    Equate the real and imaginary parts on each side:

    (2+i) x + (-3i) y = 3+i: Solve 2x = 3 and x - 3y = 1.
    I don't think this question is solved individually or am I wrong? I believe the question wants to find x and y for which supports both equations.
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  6. #6
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    Quote Originally Posted by ronaldo_07 View Post
    I don't think this question is solved individually or am I wrong? I believe the question wants to find x and y for which supports both equations.
    Mr Fantastic supposed that x and y are real, which is usually the case for unknowns called x and y.

    If x and y are complex then you can use substitution to calculate them. But you need of course both equations.
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  7. #7
    Member ronaldo_07's Avatar
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    I am still unsure as to how to solve this.

    If I put the real and imaginary numbers together I would get

    2x=3 x-3y=1
    3y=3 3x-2y=-3

    But I do not see how I can solve this with these solutions.
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  8. #8
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    Quote Originally Posted by ronaldo_07 View Post
    I am still unsure as to how to solve this.

    If I put the real and imaginary numbers together I would get

    2x=3 x-3y=1
    3y=3 3x-2y=-3

    But I do not see how I can solve this with these solutions.
    I understood them as two seperate problems and I did the first. Implicit in that is that x and y are both real (which is the usual convention).

    If they are meant to be solved simultaneously then x and y are not real. You need to provide clarification: are they two seperate questions (x and y are real) or a single question (x and y are not real).
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  9. #9
    Member ronaldo_07's Avatar
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    Both equations are one It is a "system of linear equations" I think you use matrices to solve but im not sure how to when it has complex numbers.
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  10. #10
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    Quote Originally Posted by ronaldo_07 View Post
    Both equations are one It is a "system of linear equations" I think you use matrices to solve but im not sure how to when it has complex numbers.
    Do it in the same way as you would if both equations were real. Where are you stuck using that approach?
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  11. #11
    Member ronaldo_07's Avatar
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    Im not sure how to do it as if it was real as I dont know what to do when there is i values
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  12. #12
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    Quote Originally Posted by ronaldo_07 View Post
    Im not sure how to do it as if it was real as I dont know what to do when there is i values
    Get the inverse of the 2x2 coefficient matrix in the usual way. If you understand the arithmetic of complex numbers there is nothing difficult here. Have you actually tried working through the process? Post your working of this and state where you get stuck.
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  13. #13
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    \left[\begin{array} {cc} 2+i&-3i \\ 3i&3-2i \end{array}\right]<br />
\left[\begin{array} {cc} x \\ y \end{array}\right]=<br />
\left[\begin{array} {cc} 3+i\\3-3i \end{array}\right]

    Is this correct?
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  14. #14
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    Quote Originally Posted by ronaldo_07 View Post
    \left[\begin{array} {cc} 2+i&-3i \\ 3i&3-2i \end{array}\right]<br />
\left[\begin{array} {cc} x \\ y \end{array}\right]=<br />
\left[\begin{array} {cc} 3+i\\3-3i \end{array}\right]

    Is this correct?
    Yes. Now calculate the inverse matrix. Then left multiply both sides by the inverse matrix. Then read off the answer for x and y.
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  15. #15
    Member ronaldo_07's Avatar
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    \left[\begin{array} {cc} 2+i&-3i \\ 3i&3-2i \end{array}\right]<br />
\left[\begin{array} {cc} x \\ y \end{array}\right]=<br />
\left[\begin{array} {cc} 3+i\\3-3i \end{array}\right]

    \left[\begin{array} {cc} 2+i&-3i \\ 3i&3-2i \end{array}\right]^{-1} \left[\begin{array} {cc} 3+i\\3-3i \end{array}\right]= \left[\begin{array} {cc} x \\ y \end{array}\right]

    \frac{1}{17-i}\left[\begin{array} {cc} 3-2i&3i \\ -3i&2+i \end{array}\right]=\left[\begin{array} {cc} x \\ y \end{array}\right]

    After this it gets really messy and I am unable to get a solution.
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