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Application of the identity matrix II. v05

Submitted By: xhunga
Rating: starstar (Rate It)


You have seen :
 
 a)  You must multiply m by four identity matrix,
     to eliminate all the values below the pivots.

       11/1 14/1 -16/1  19/1
  m =  12/1 13/1  13/1  11/1
        0/1  0/1  17/1  11/1
       12/1 18/1   1/1 -16/1

  mID1 : to eliminate all the values under the first  pivot.
  mID2 : to eliminate all the values under the second pivot.
  mID3 : to eliminate all the values under the third  pivot.
  mID4 : to eliminate the pivot.

                            Can you do that in one step ?
                           ===========================

    Multiply  mID4 by mID3 by mID2 by mID1, in this order.

         mID = ((mID4 * mID3) * mID2)* mID1.

               And then mutiply mID by m. 

                1/1 14/11 -16/11  19/11 
    mID * m  =  0/1  1/1  -67/5  107/25
                0/1  0/1    1/1   11/17
                0/1  0/1    0/1    1/1

   All the pivots equal 1 (1/1).
   All the values below the pivot equal 0 (0/1).

   The work must be do without swap two rows.
   -----------------------------------------

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