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System of equations

  1. May 4, 2016 #1
    1. The problem statement, all variables and given/known data
    Okay this is the problem it seems so easy but i just cannot for the life of me get it to click into my mind,

    I have 4 unknowns and 5 equations and i have to put it into a matrix and try solve it matricies or eigenvalues/eigenvectors.

    The 5 equations are:

    a= b/2
    b=a/3 + d

    2. Relevant equations

    I know i can attempt the question via system of linear equations and Ax=b
    3. The attempt at a solution

    Okay so this follows on from a earlier sections of the question. These 4 equations are derived and the last one is stated.

    If you make a matrix as so:

    0 1/2 0 0
    1/3 0 0 1
    1/3 1/2 0 0
    1/3 0 1 0
    1 1 1 1

    Am I not sure what to make b equal
    I know x is simply [ a b c d]

    Any help would be appreciated :)
  2. jcsd
  3. May 4, 2016 #2


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    Science Advisor

    Since you have five equations and only four unknowns, the system is overdetermnied and might not have any solutions. It turns out that one of the first four equations is redundant, so there is a unique solution.

    If you just want a solution, this is easily solved by just eliminating variables. For example, use the first equation to write b in terms of a, the third equation to write c in terms of a, and the fourth equation to write d in terms of a, then plug them all into the last equation to find a.

    But assuming you want to use matrix methods, you first need to eliminate one of the equations, then write them all in a form with all of a,b,c,d on the left hand side, and any constants on the right hand side. For example, If you throw out the fourth equation, you would write them as:

    a-b/2 = 0
    -a/3+b-d = 0
    -a/3-b/2+c = 0
    a+b+c+d = 1

    Then you can write this as AX = B, with X = (a,b,c,d) and B = (0,0,0,1). You can then solve for X, and plug the results in to the fourth equation you threw out to see if the solution works in this equation (which it does). Does this help?
  4. May 4, 2016 #3
    Yes that does help a little bit. So you are saying if i late out a matrix like this:

    1 -1/2 0 0 0
    -1/3 1 0 -1 0
    -1/3 -1/2 1 0 inverse * 0
    1 1 1 1 1

    i can solve?

    Update: I can eliminate equation 4

    But how can i eliminate it?
  5. May 4, 2016 #4


    Staff: Mentor

    Not the way you have it. Your matrix is missing a row for one of the equations, and you can't just throw one equation away. As phyzguy already said, your system of equations has 5 equations but only 4 variables, which makes it an overdetermined system.
    Also, in the future, please make more of an effort to give your threads a meaningful title. "Absolutely bugging me" doesn't give a reader any idea of the nature of the problem.
  6. May 4, 2016 #5

    Sorry about the title.

    Okay that I know what A equals and that b equals (0,0,0,1). Would I be able to do A^-1 * b to get the solutions of X?
  7. May 4, 2016 #6


    Staff: Mentor

    b does NOT equal (0, 0, 0, 1) -- b is a number.
    Your matrix A is 5 X 4, meaning it has 5 rows with 4 columns. It's not a square matrix. A matrix inverse is possible only for square matrices, and not all square matrices have inverses.
  8. May 4, 2016 #7


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    Staff Emeritus
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    Homework Helper

    In order to make a proper matrix equation for the system above, rewrite the 5 equations as follows:

    a - b/2 = 0
    a/3 -b + d = 0
    a/3+b/2 - c = 0
    a/3+c - d = 0

    Now, your matrix equation becomes:

    $$\begin {bmatrix}\
    1 & -1/2 & 0 & 0\\
    1/3 & -1 & 0 & 1\\
    1/3 & 1/2 & -1 & 0 \\
    1/3 & 0 & 1 & -1 \\
    1 & 1 &1 & 1
    \end {bmatrix} \times \left [ \begin {array} {c} a \\ b\\ c \\ d \end {array} \right ] = \left [ \begin {array} {c} 0 \\ 0 \\ 0 \\ 0 \\ 1 \end {array} \right ] $$
  9. May 4, 2016 #8

    Ray Vickson

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    Using matrix methods in this problem is a waste of time and works against your understanding in this case. Just using elementary algebra to solve for b, c, d in terms of a (from the first three equations) can be done faster and easier than you could even write out the matrix.

    Anyway, a crucial point is that if (a,b,c,d) satisfies the first four equations, then so does (ka,kb,kc,kd) for any real constant k, so the first four equations cannot determine the variables uniquely. The fifth equation is crucial towards the process of getting a solution, if a solution exists at all.

    So: use the first three equations to express b,c,d in terms of a; plug those expressions into the fourth equation to check if the equations are consistent. If the fourth equation does not work, you are done: the system has no solution. If the fourth equation does work, then use the fifth equation to finally determine the missing value of a.
    Last edited: May 4, 2016
  10. May 4, 2016 #9


    Staff: Mentor

    I'm pretty amazed that you can talk about solving this with matrix methods "or eigenvalues/eigenvectors" and then write "b equals (0, 0, 0, 1)." In any case, this problem has nothing to do with eigenvalues or eigenvectors.

    I second Ray's suggestion of NOT using matrices, but instead work directly with the system of equations, to promote your understanding what's going on in this problem.
  11. May 5, 2016 #10


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    There is a lot of confusion in this thread.

    (1) First of all, several people have confused the b in (a,b,c,d) with the b in the matrix equation a.x = b. That is why I used capital letters for the matrix equation in Post #2. So let's agree to write the matrix equation as A.X = B. Then there is no confusion.

    (2) Second, in general, only square matrices have inverses. You can talk about left and right inverses of non-square matrices, but since it sounds like you are just learning linear algebra, that's probably more complexity that you want at this point.

    (3) Third, while I agree that the problem is easily solved with elimination and there is no need to use matrices, you can get a solution using matrices if you want. But to do this, you have to eliminate one of the equations so that you have a square matrix. You can eliminate one equation by simply ignoring it. Then you find a solution to the remaining four equations, and then plug that in to the equation you ignored. If the equation is satisfied, you have found a solution to all five equations. If it isn't, then there is no solution. You can try it several ways by changing which equation you eliminate. For example, if you eliminate the last equation, you will find that the matrix is singular and has no inverse. If you eliminate the fourth equation, then the matrix equation (A.X = B) becomes:
    \end{bmatrix} [/tex]

    This can be solved using X = A^-1 . B, and this solution also solves equation 4 that you eliminated. The solution is also the same one you arrive it by simple elimination. Does this make sense?
  12. May 31, 2016 #11

    Yes it made much more sense, I eventually worked it out. Was a lot easier than I thought!
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