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I Finding the inverse of a matrix using transformations?

  1. Jun 26, 2017 #1
    We use A = I.A as equation and then by transforming only A of LHS and I of RHS we come to I = P.A and we say that P is the inverse of matrix A
    My question is that why we only tranform A and I, why A of RHS is left as it is during the transformation, or why transformation do not take place in both the part of RHS
     
  2. jcsd
  3. Jun 26, 2017 #2

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    Each transformation step, T, is multiplication on the left: TA = TIA. It is not also multiplying the A of the RHS. TA ≠ TITA
     
  4. Jun 26, 2017 #3
    Can you prove it
    Or is it like i am transforming only one matrix on LHS therefore i have to transform only one from RHS
     
  5. Jun 26, 2017 #4

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    You must do the same things on both sides to keep them equal. Left multiply both sides to get TI = TIA. That keeps the two sides equal. Given a series of transformations Tn...T2T1, You have Tn...T2T1I = Tn...T2T1IA.

    You should not "double up" on the right-side transformations like Tn...T2T1ITn...T2T1A
     
    Last edited: Jun 26, 2017
  6. Jun 26, 2017 #5
    Is their a proof which says that if A =BC
    Then the transformation of A is done simultaneously with transformation of B or C but not with both, i know that first one is right but is their a proof
     
  7. Jun 26, 2017 #6

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    Matrix multiplication is similar to multiplication of real numbers, except that it is not commutative and there is not a guaranteed multiplicative inverse. This should be proven early in the theory of matrices. So this question is very similar to a question about real numbers: Since 6 = 2×3, is 4×6 = 4×2×3 or should it be 4×6 = 4×2×4×3

    PS. Studying the similarity and differences of linear transformations, matrices, real numbers, etc. is a good motivation for the study of abstract algebra. That is where these issues are addressed directly.
     
  8. Jun 27, 2017 #7
    IMG_20170627_212308393_HDR.jpg
    IMG_20170627_212329829_HDR.jpg
    IMG_20170627_212345731_HDR.jpg
    I got this proof from an old book but its really satisfying
     
  9. Jun 27, 2017 #8

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    Good proof. One way to quickly see that your alternative hypothesis ( TA = TITA ) was wrong is to realize that the real numbers are 1x1 matrices. So a counterexample in the reals is also a counterexample in matrices. That can sometimes allow you avoid trying to prove something that is wrong. Of course, there are things that are true for the reals that are not true for general matrices, so checking something in the real numbers won't always tell you if something is wrong for matrices.
     
  10. Jun 27, 2017 #9
    Actually my focus was in poving the general case instid of checking the result
     
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