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Find a certain 3x3 matrix

  1. Oct 20, 2004 #1
    Hello everyone the following problem has me completely stumped, I am to find a certain 3x3 matrix D that satisfies the following equation:

    [tex]ADA^{-1}[/tex] = [tex]\left(\begin{array}{ccc}1&0&0\\1&0&0\\1&0&0\end{array}\right)[/tex]

    where :

    [tex]A = \left(\begin{array}{ccc}1&2&3\\0&1&1\\0&2&1\end{array}\right)[/tex]

    [tex]A^{-1} = \left(\begin{array}{ccc}1&-4&1\\0&-1&1\\0&2&-1\end{array}\right)[/tex]

    Heres my reasoning (or lack thereof), I know that [tex]AA^{-1}[/tex] will yield the identity matrix I3, however clearly the D im looking for is WITHIN this operation, and by matrix multiplication i cannot use this fact since the order is now completely different. But what I do know is how to find the inverse of A, but what property can I use for finding a 3x3 matrix? You see this would be simpler if they were happening to look for a 3x1 matrix D where I could use row operations in gauss jordan elimination to solve for the particular values, however I did not find any examples of this problem in the book--- where I am given an unknown nxn matrix to find and a certain operation that it must adhere to.

    I could i solve this one? I have been understanding everything up to this point but i am clearly not understanding some simple rule--- thanks a lot for your help.
    Last edited: Oct 20, 2004
  2. jcsd
  3. Oct 20, 2004 #2
    Multiply both sides on the left by A inverse, then multiply both sides on the right by A...then D is on the left and you can expand out the other to find what D is.
    Last edited: Oct 20, 2004
  4. Oct 20, 2004 #3
    Wait what do you mean by expanding out the other? What does both sides on the left mean? Like A and D? Thanks.
  5. Oct 20, 2004 #4
    btw i really apologize for the stupid thread title... i was ctually testing out my TeX format and accidentally posted with a wrong name--- id change it if i could but i cannot!!
  6. Oct 20, 2004 #5
    lol...its ok :)

    EDIT: changed to tex







    Last edited: Oct 20, 2004
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