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Solutions of matrices

  1. Dec 31, 2003 #1
    I'm trying to learn linear algebra by myself from a book called "Introduction to linear algebra" by A.D. Martin and V.J. Mizel. One point I'm so far pretty confused about is whether a matix has a solution only if m equals n? I think the book says that if m < n the matrix has infinite solutions, which makes sense, but it doesn't say anything about when m > n. In that case, is there a solution?

    The book has problems for you to solve, but no answers. That doesn't matter if a matrix has a solution you can verify, but I'm getting a suspicious number of matrices that have no solutions. I think I don't understand the Gaussian reduction algorithm well enough, at least I find that the following matrix has no solutions, when according to the book it should since m = n.

    I'll write it as an equation, I have no idea how to do it properly in latex.

    2x + 3y + z = 5
    x + 0y - z = 1
    2x - 9y - 11z = -5
     
    Last edited: Dec 31, 2003
  2. jcsd
  3. Dec 31, 2003 #2
    m? n? You might want to clarify what you mean by those variables...

    I'll "do" your example.

    2x + 3y + z = 5
    x + 0y - z = 1
    2x - 9y - 11z = -5

    Matrix form:

    Code (Text):

    2 3 1 | 5
    1 0 -1 | 1
    2 -9 -11 | -5
     
    Move the middle row:

    Code (Text):

    1 0 -1 | 1
    2 3 1 | 5
    2 -9 -11 | -5
     
    Add the (new) first row to the middle row:

    Code (Text):

    1 0 -1 | 1
    3 3 0 | 6
    2 -9 -11 | -5
     
    Add -11 times the first rwo to the last row:

    Code (Text):

    1 0 -1 | 1
    3 3 0 | 6
    -9 -9 0 | -16
     
    Divide the middle row by 3, and the last row by -9:

    Code (Text):

    1 0 -1 | 1
    1 1 0 | 2
    1 1 0 | 16/9
     
    Do you see the conflict between the middle and last row? You want x + y = 2 /and/ x + y = 16/9, and surely no such numbers can exist, hence the system has no solutions. The book must be mistaken.
     
    Last edited: Dec 31, 2003
  4. Dec 31, 2003 #3

    ahrkron

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    Staff Emeritus
    Gold Member

    Think about it geometrically. Say, with two variables, each equation corresponds to a straight line. With only two equations, the only point where they intersect is the solution. If you add one extra equation, it is possible that the extra line also contains that point (in which case the common intersection point is the overall solution), but it may not pass through it; if so, the system can be said to be inconsistent: If you choose two out of the three equations, you'll have a solution, but they will be different for each pair.
     
  5. Dec 31, 2003 #4
    Re: Re: Solutions of matrices

    I see, that makes sense. Thanks.

    Muzza, I'm sorry, I should have said what I mean by m and n. I mean the rows and columns of the matrix. Mathworld.com uses the same notation so I assumed it was universal. But I also got that the matrix had no solutions, I just thought I didn't understand the algorithm enough and had made a mistake somewhere. Thanks for the help.

    This stuff is way deeper than I thought. If I understand this stuff correctly a matrix only has a solution if the number of rows equal the number of columns, but even then it might not.
     
  6. Dec 31, 2003 #5
    By "a solution", do you mean "a unique solution"? If so, you're basically correct. The "might not" scenario (I believe) happens when one (or more) of the equations is a linear combination of the others, then it has an infinitude of solutions (or when the system "has a contradiction" in it, such as your example, then it has no solutions).

    Btw, by "the matrix", do you mean just the coefficient matrix or the augmented matrix?
     
    Last edited: Dec 31, 2003
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