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Since, I can't divide vectors to deduce an inverse matrix I have dismissed that approach. I did find that if I multiply all of the matrix row operators It will yield the inverse. Since I did the logic work to put my original matrix into RRef. I can use this approach. Problem is, I don't understand how to place the multipliers in the identity matrix to reflect the logic of the RRef process. Can any one help me , or post a link?

Chris
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  • Found a link to the Gauss/Jordan method. Let me see if I have this right. I set up the matrix just like I did for using RRef to find solution of a system of equations only I augment the Coordinate matrix with the Identity matrix and then I use the same RRef solution process. Am I understanding this right? – Chris Nov 04 '13 at 14:25
  • Do you have to find the inverse through Gauss method, or are you allowed to find through different linear algebra means? I am not talking about punching it into the calculator... – imranfat Nov 04 '13 at 14:43
  • This is to add things to my own back of tricks. I am trying to find AND understand the easiest way to do it. Gauss will let me find the inverse w/o screwing around with finding the determinant that gives it one thumbs up from me. – Chris Nov 04 '13 at 14:49
  • Are you familiar with the Characteristic equation and Cayley Hamilton Theorem for matrices? – imranfat Nov 04 '13 at 14:52
  • No I am not. But one thing I have learned recently is, the more advanced the math the simpler the solution. Well at least when you understand the process. So this sound really cool. Care to give a tinkerer a lesson? – Chris Nov 04 '13 at 14:54
  • It's too much to go into detail about this if you have not studied eigenvalues and eigenvectors yet. Once you learn about those, you have learned how to set up the Characteristic equation. Cayley Hamilton essentially states that the matrix satifies its own characteristic equation. From that, the inverse can be expressed as a linear combination of the matrix itself and the identity matrix. It becomes real easy then. Here is a link: http://mathforum.org/library/drmath/view/51978.html – imranfat Nov 04 '13 at 16:00
  • Thanks for the link... It going to take so time to figure it out but looks interesting. – Chris Nov 05 '13 at 02:55

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