Update 2013

3.3-8   A:=Matrix([[1,-4,-5],[3,-9,3],[1,-2,3]]); Find rref(A).
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 A matrix is in reduced row echelon form <==> the corresponding
 homogeneous system of equations is a reduced echelon system. There is a
 simple criterion for testing if a given matrix B is in reduced row
 echelon form:

   TEST: Each nonzero row of B has a leading one, and the leading one
         sits in a column of the identity. Finally, these identity
         columns appear in natural order as consecutive initial columns
         of the identity.

 The textbook reports the answer as the 3x3 identity matrix. Details
 expected include a frame sequence that starts with matrix A and ends
 with the identity. Document all steps with detailed information in the
 form swap(s,t), combo(s,t,c), mult(t,m).


3.3-18  A:=Matrix([[1,-2,-5,-12,1],[2,3,18,11,9],[2,5,26,21,11]]);
=====   Find rref(A)
 The textbook reports the correct answer as a 3x5 matrix B,
 
       B=rref(A)=Matrix([[1,0,3,-2,3],[0,1,4,5,1],[0,0,0,0,0]])

 Details expected include a frame sequence that starts with matrix A and
 ends with matrix B. Document all steps with detailed information in the
 form swap(s,t), combo(s,t,c), mult(t,m).

 The matrix B is a reduced row echelon form, because the nonzero rows
 are rows 1 and 2, and each contains a leading one, which sits in a
 column of the 3x3 identity matrix I. These columns are the first two
 columns of I in natural order.

 MAPLE ANSWER CHECK.
       A:=Matrix([[1,-2,-5,-12,1],[2,3,18,11,9],[2,5,26,21,11]]);
       linalg[rref](A);

3.3-20   A:=Matrix([[3,6,1,7,13],[5,10,8,18,47],[2,4,5,9,26]]);
======   Find rref(A).
 The textbook claims the correct answer is a 3x5 matrix B,
 
       B=rref(A)=Matrix([[1,2,0,2,3],[0,0,1,1,4],[0,0,0,0,0]])

 Details expected include a frame sequence that starts with matrix A and
 ends with matrix B. Document all steps with detailed information in the
 form swap(s,t), combo(s,t,c), mult(t,m).

 The matrix B is a reduced row echelon form, because the nonzero rows
 are rows 1 and 2, and each contains a leading one, which sits in a
 column of the 3x3 identity matrix I. These columns are the first two
 columns of I in natural order.

 MAPLE ANSWER CHECK.
       A:=Matrix([[3,6,1,7,13],[5,10,8,18,47],[2,4,5,9,26]]);
       linalg[rref](A);

3.3-34
======
 The 3x3 matrix forms possible for a reduced echelon form depends on the
 number of leading ones.

 Case 1. Three leading ones.

   Then A:=Matrix([[1,*,*],[0,1,*],[0,0,1]]) initially
   but to be an rref there has to be zeros above and below a leading
   one. This forces the form

        A:=Matrix([[1,0,0],[0,1,0],[0,0,1]])

 Case 2. Two leading ones.

   Then A:=Matrix([[1,*,*],[0,1,*],[0,0,0]]) initially
   but to be an rref there has to be zeros above and below a leading
   one. This forces the form

        A:=Matrix([[1,0,*],[0,1,*],[0,0,0]])

 Case 3. One leading one.

   Then A:=Matrix([[1,*,*],[0,0,0],[0,0,0]]) initially, but
   there are nochanges to be made, giving

        A:=Matrix([[1,*,*],[0,0,0],[0,0,0]])

 Case 4. No leading one.

   The A has to be the zero matrix (otherwise, one entry could be made
   into a leading one).

        A:=Matrix([[0,0,0],[0,0,0],[0,0,0]])

==============