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Inverse of a matrix

  1. Feb 16, 2016 #1
    1. The problem statement, all variables and given/known data
    a &b \\

    I'm supposed to find the inverse

    2. Relevant equations

    Method of Gauss-Jordan
    3. The attempt at a solution
    So I tried putting zeros in this and I got the following :

    ad-ac &0 &ad &ad-a \\
    0&bc-ad &c &-a

    Does my procedure seem to make any sense ?

    I know that after that I must have two "1"
  2. jcsd
  3. Feb 16, 2016 #2


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    Where is your inverse matrix? It should be 2x2.
    Just multiply it with the original matrix to check that you got the right result.
  4. Feb 16, 2016 #3


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    No, what you're doing doesn't make any sense.

    the definition of the inverse of matrix A is

    A * A-1 = I,

    where A-1 is the inverse of A and I is the identity matrix, which contains 1 on the main diagonal and is zero everywhere else.

    You can calculate the inverse by solving the system of equations above for multiple RHS as indicated:

    Code (Text):

    | a   c | | x1  x2 |   | 1   0 |
    | b   d | | x3  x4 | = | 0   1 |
    The system above can be solved by Gaussian elimination. The values x1 ... x4 will be the elements of the inverse matrix.

  5. Feb 16, 2016 #4
    I was finally able to solve the problem. Thank you anyway for your answer. Im still stuck on the Gaussian elimination of the other problem though :/
  6. Feb 16, 2016 #5


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    No, what you have shown there does NOT make any sense! Where did you get it?
    It is true that one method of finding the inverse to a given matrix is to write the given matrix with the identity matrix beside it:
    [tex]\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}[/tex]
    then use "row operations" to reduce the given matrix to the identity matrix while at the same time applying those same row operations to the identity matrix.
    A good way to do the row operations is to work one row and column at a time. For example, The first few step should be to divide the entire first row by a, in order to get "1" there, then subtract that new first row, times c, from the second row, to get "0" there:
    [tex]\begin{bmatrix}1 & \frac{b}{a} \\ 0 & d- \frac{bc}{a}\end{matrix}\begin{bmatrix}\frac{1}{a} & 0 \\ -\frac{c}{a} & 1 \end{bmatrix}[/tex]

    Now, divide the second row, of both matrices, by [tex]d- \frac{ac}{b}= \frac{ad- bc}{b}[/tex] to get a "1" in the last place (you may recognize that "db- ac" as the determinant of the original matrix) then subtract that times the b/a from the first row- again for both matrices. That will give
    [tex]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix}\frac{d}{ad- bc} & \frac{-b}{ad- bc} \\ \frac{c}{ad- bc} & \frac{a}{ad- bc} \end{bmatrix}[/tex]
  7. Feb 16, 2016 #6


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    I think what the OP was doing was setting up the augmented matrix
    $$\left[\begin{array}{cc|cc} a & b & 1 & 0 \\ c & d & 0 & 1\end{array}\right]$$ and then using Gaussian elimination to turn the lefthand side into the identity matrix. The righthand side would then be the inverse. The method is fine, but the fact that the OP's matrix has no ##b## on the righthand side indicates something went wrong.
  8. Feb 16, 2016 #7
    I solved the problem, there's no need now.
  9. Feb 16, 2016 #8


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    It's not all about you. ;) I was clarifying your unclear and confusing original post for the benefit of others, like SteamKing.
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