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Solve system of equations with inverse matrix

  1. May 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the inverse matrix of A, then use this inverse to solve system of equation.

    A is a given 3 x 3 matrix and the system of equations is 3 equations in 3 unknowns.

    2. Relevant equations



    3. The attempt at a solution

    I have found the inverse of A using an augmented identity matrix. Checked this inverse by taking product of A and inverse A and got identity matrix.

    The system of equations is something like this (these are not the actual values):

    6x - 2y = 10
    14x - 4y + 4z = 3
    6x - 2y + 2z = 14

    Now when I take the product of inverse A and [10 3 14] to solve for x, y and z, putting these values into the original equations does not hold.

    I was told to try re-arrange the equations and compare to the inverse matrix, but can't seem to find a solution.

    Note that by swapping 2 rows in my inverse matrix the co-effiecients of the unkowns in the system of equations are the same as the entries in the matrix.

    Gee I hope that made sense.
     
  2. jcsd
  3. May 19, 2009 #2

    Mark44

    Staff: Mentor

    Without knowing what the system of equations is, it's not possible to know whether the system is inconsistent or you have made a mistake. If you have matrix equation A[x y z]T = [c1 c2 c3]T, and A has an inverse, then the solution to the matrix equation is [x y z]T = A-1[c1 c2 c3]T.
     
  4. May 19, 2009 #3

    Cyosis

    User Avatar
    Homework Helper

    It would be helpful if you showed us what you took as A and whether or not you inverted it correctly.

    To repeat what Mark said in a slightly less abstract manner, this is the system you're trying to solve:
    [tex]
    \left( \begin{matrix} 6 & -2 & 0 \\ 14 & -4 & 4 \\ 6 & -2 & 2 \end{matrix}\right) \left( \begin{matrix} x \\ y \\ z \end{matrix}\right) =\left( \begin{matrix} 10 \\ 3 \\14\end{matrix}\right)
    [/tex]

    Did you take the matrix on the left hand side as A?
     
    Last edited: May 19, 2009
  5. May 19, 2009 #4
    The system is consistent, I can solve it using row operations, however we are explicitly asked to use the inverse matrix of A to solve this particular system.

    I am given this matrix A:
    [tex]
    A = \left( \begin{matrix} 1 & 0 & -2 \\ 3 & 1 & -6 \\ 0 & 1 & 1 \end{matrix}\right)
    [/tex]

    And produce this inverse:
    [tex]A^{-1} = \left( \begin{matrix} 7 & -2 & 2 \\ -3 & 1 & 0 \\ 3 & -1 & 1 \end{matrix}\right) [/tex]

    now I must use this inverse to solve the system. This is a 2 part question, each part with a system to solve. Part a works with this inverse. Part b (the system given in this thread) does not. The actual given system is:

    3x - y = 5
    7x - 2y + 2z = 3
    3x - y + z = 7

    Using row operations I get to row-echelon form and use back substitution to get z = 2, y = -38 and x = -11 and the original equations hold.

    Using the inverse matrix gives z = 19, y = -12 and x = 43 and using these the original equations do not hold. I do this:

    [tex]\left( \begin{matrix} x & y & z \end{matrix}\right) = \left( \begin{matrix} 7 & -2 & 2 \\ -3 & 1 & 0 \\ 3 & -1 & 1 \end{matrix}\right) \left( \begin{matrix} 5 & 3 & 7 \\ \end{matrix}\right)[/tex]

    The co-efficients in the system are almost the same as the entries in the inverse matrix.
     
    Last edited: May 19, 2009
  6. May 19, 2009 #5

    rock.freak667

    User Avatar
    Homework Helper

    3x - y = 5
    7x - 2y + 2z = 3
    3x - y + z = 7

    Re-write this as

    7x - 2y + 2z = 3
    -3x + y = -5
    3x-y+z=7

    Now write down this in the matrix for AX=B.

    Take a look at the matrix A and the inverse you calculated before.
     
  7. May 20, 2009 #6
    The entries are the same, but then? Perhaps I am missing some idea or principle.

    If AX = B, then A is the co-efficients, B is the numbers after the equals, X is the vector (x y z) but where does the inverse come into the story?
     
  8. May 20, 2009 #7
    rock.freak667 had the solution.

    3x - y = 5
    7x - 2y + 2z = 3
    3x - y + z = 7

    Re-write this as

    7x - 2y + 2z = 3
    -3x + y = -5
    3x-y+z=7

    By letting B = the co-efficients of the system then B = inverse A, so A = inverse B.

    then, Bx = b, where b = [3 -5 7]
    so, x = inverse B b
    thus x = A b

    solving this for x, y and z and checking original system holds. YES!!!

    Thanks to everyone for the help.
     
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