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Inverse of a 2X2 matrix

  1. Sep 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the inverse of matrix A

    02
    30

    3. The attempt at a solution

    I was thinking of doing a row swap to get a diagonal matrix with nonzero diagonal entries, PA (a.k.a. B). I want this matrix's inverse, B-inverse (easily found by dividing the ones of the identity matrix by the diagonal entries) to serve as a means to get to A-inverse

    I want to use this relationship specifically: (B-inverse)(PA)=(A-inverse)(A)=I. I want to multiply all sides by A-inverse to show that (B-inverse)(P)=(A-inverse), but I am really shaky as to how I properly utilize the multiplication rules for matrices in this case.

    For example: Would multiplying both sides by A-inverse cancel out A? Wouldn't I be applying A-inverse to the outermost matrix and not even hit A?

    To sum: Where I really get lost is how to properly manipulate matrix equations in order to cancel.

    Let me know if you have any questions. Sorry I couldn't make the post more visual. Don't know how to draw out matrices.
     
  2. jcsd
  3. Sep 13, 2010 #2
    You could just use the identity that any matrix times its inverse equals the identity matrix.
     
  4. Sep 13, 2010 #3
    To use this identity you augment your matrix with the identity matrix. So you start with (A|I) where I is the identity matrix. You use matrix operations to transform A into I and your augmented matrix will be A^-1.

    So you start with:

    02|10
    30|01

    And you want to use elementary row operations to make your matrix to look like:

    10|ab
    01|cd

    Where a,b,c,d will make up you’re a^-1
     
  5. Sep 13, 2010 #4
    This is Guass-Jordan, no? So even though I am permutating the original matrix to get another matrix, the end result is A^-1, not the inverse of the matrix after permutation?
     
  6. Sep 13, 2010 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Or just use [tex]\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}0 & 2 \\ 3 & 0\end{bmatrix}= \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}[/tex]
    to get the equations 0a+ 3b= 3b= 1, 2a+ 0b= 2a= 0, 0c+ 3d= 3d= 0, and 2c+ 0d= 2c= 1. Those equations are pretty close to begin trivial, aren't they?
     
  7. Sep 13, 2010 #6

    Mark44

    Staff: Mentor

    There's a formula that can be used to find the inverse of a 2x2 (only) matrix.
    [tex]\text{If} A = \begin{bmatrix}a & b \\ c & d\end{bmatrix},[/tex]
    [tex]A^{-1} = \frac{1}{|A|}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}[/tex]

    Of course, it must be true that |A| is not 0.
     
  8. Sep 13, 2010 #7
    At the moment we are ignoring the determinant formula for the 2x2 case. That's what that is, right, Mark 44?

    Thanks for all the help. I don't know why I was going the direction I was with this. For some reason I keep thinking this class requires a ridiculously abstract approach when, in most cases, I can use simpler mechanisms.
     
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