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Row operations performed on two matrices

  1. Jun 5, 2012 #1
    if you perform row operations on a matrix A to convert it to the identity matrix and then use the same row operations and apply it to another matrix B, why is it that the end result of B^r does not depends on B's actual sequence
     
  2. jcsd
  3. Jun 5, 2012 #2
    What do you mean by B's actual sequence?
     
  4. Jun 5, 2012 #3

    HallsofIvy

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    And, what do you mean by "B^r"? Every row reduction is equivalent to an "elementary matrix"- the result of applying that row reduction to the identity matrix. Applying a given row operation to a matrix is the same as multiplying the corresponding elementary matrix. And applying row operations to A to reduce it the identity matrix means that the product of the corresponding elementary matrices is [itex]A^{-1}[/itex]. Applying those row operations to B gives [itex]A^{-1}B[/itex].

    That means, in particular, that if you have the matrix equation Ax= B, and apply the the row operations that reduce A to the identity matrix to B, you get [itex]x= A^{-1}B[/itex], the solution to the equation.
     
  5. Jun 5, 2012 #4
    When I say Bs actual sequence, I mean the numbers that compose that matrix such as a 3x3 matrix with the numbers 654,896,327 and when I say Br I mean performing the exact same row operations that you did on A and applying them to B in the same order and I want to know why it doesn't matter what the actual sequence of B is as long as you're performing the same row operations on it as you did with another matrix, A
     
  6. Jun 5, 2012 #5
    I guess the short answer is that the result you get does depend on the entries of B in exactly the way that HallsofIvy explained.

    What doesn't matter I guess is the exact sequence of steps you took to row reduce A. As long as you do row operations that eventually reduce A to the identity, the result of all those row operations combines to be the same operation. When you apply that operation on B, you'll always get the matrix A^(-1)B.
     
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