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S2013 notes on problems 3.1-4,6,8,12,16,18,26
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3.1-4  5x-6y=1, 6x-5y=10
=====
 The answer "x=5, y=4" is the unique solution, as reported by the
 book. Submitted work is expected to display the solution details,
 which include documented toolkit steps of swap, combo, mult. Please
 do not display arithmetic steps. Preferred is a frame sequence
 which ends with last frame x=5, y=4. Each frame uses exactly one of
 the 3 rules: combo, swap, mult.

3.1-6  4x-2y=4, 6x-3y=10
=====
 The answer is "no solution" because of a signal equation, obtained
 with the toolkit of 3 rules: combo, swap and mult. A signal
 equation is a nonzero equation having no variables, like "0=1".

 Please show the frame sequence details, one operation per display.
 Point out the signal equation in the last display.

3.1-8  3x-6y=12, 2x-4y=8
=====

 Here's a sample of what should be displayed in a solution, using
 frame sequences and the operations swap, combo, mult. 
 
 The term "Frame" refers to a video frame, as though the document
 written was videoed, and only the relevant frames were extracted
 for view.

   3x-6y=12,  Frame 1, original problem
   2x-4y=8

   x-2y=4,    Frame 2, mult(row=1,multiplier=1/3)
   2x-4y=8

   x-2y=4,    Frame 3, 
   0x+0y=0    combo(src=1,target=2,multiplier=-2)

   x-2y=4,    Frame 4, last frame, reduced echelon form
      0=0

 The last frame algorithm is applied, because there is one lead
 variable (x) and one free variable (y). Assign invented symbol t1
 to the free variable, then back-substitute until all variables are
 expressed in terms of the invented symbols. Then list the
 variables, in variable list order, so expressed, as the answer or
 general solution.

  x = 4+2y=4+2t1,    Final general     x = 4+2t1,
  y = t1             solution is       y = t1.

3.1-12     3x+5y- z=13,   Solve for the general solution,
======     2x+7y+ z=28,   if it exists, otherwise explain
            x+7y+2z=32    why there is no solution.

  The details will be described for equations only, because the
  matrix interface does not happen until later in the textbook.
  However, this exercise will likely be solved after a week of
  exposure to linear algebra and matrices, hence the remarks below
  about maple.

  Because maple is used as an assistant, you could first program
  this system into maple and find the solution.

   MAPLE ANSWER CHECK.
     A:=Matrix([[3,5,-1],[2,7,1],[1,7,2]]);
     b:=<13,28,32>;
     linalg[linsolve](A,b);
     C:=<A|b>;        # First Frame
     linalg[rref](C); # Last Frame
     # Answer: Au=b has solution u=<1,3,5>.     
  
  DETAILS.
  The problem is to write the details by hand, for a frame sequence
  that starts at the original system and ends at the reduced row
  echelon form of C, which must be

    x=1,
    y=3,
    z=5.
  
  To start, frame 1 is the original system,
  
     3x+5y- z=13,   Frame 1
     2x+7y+ z=28,   
      x+7y+2z=32    
      
  To find frame 2, we identify the appearance of variable x in the
  system, which is the first variable in the list order x,y,z. If x
  did not appear, then we would consider y, and then z. If dealing
  with the trivial system 0=0, 0=0, 0=0, then there are no
  variables, hence no lead variables, all variables are free
  variables.

  We try to make x a lead variable. Rules for display require that
  lead variable x appear in equation 1. So we arrange this by a swap
  of equations.

      x+7y+2z=32,   Frame 2, swap(source=1,target=3)
     3x+5y- z=13,
     2x+7y+ z=28,   
      
 The plan to make x a lead variable continues. The idea is to remove 
3x from equation 2, and 2x from equation 3, using combo. This takes 
two frames, because we use only one operation per frame. Frame 4 
then looks like this:

      x+7y+2z=32,   Frame 4, swap(source=1,target=3)
      0+ * * =*,
      0+ * * =*,   

 The rest of the work proceeds systematically. Once x is a lead
 variable, then it appears only in equation 1. If y is to be a lead
 variable, then it must appear in equation 2 or 3, and we must load
 equation 2 with the only y appearing in the whole frame.

 Finally, once x,y are lead variables appearing only once each in
 the whole frame, which must be equations 1 and 2, by design rules,
 then we look for the appearance of variable z in the last remaining
 equation. If absent, then z is a free variable, if present then it
 can be made into a lead variable.

3.1-16    x-3y+2z=6, x+4y-z=4, 5x+6y+z=20  Find the general solution
======    or else explain why there is no solution.

 The answer is "no solution" because of a signal equation, obtained
 with the toolkit of 3 rules: combo, swap and mult. A signal
 equation is a nonzero equation having no variables, like "0=1".
 Because there are 3 variables x,y,z, please develop a frame
 sequence, in which each frame uses just one of the 3 rules, ending
 with a frame that has a "0=1" equation.

 MAPLE ANSWER CHECK.
  The reduced row echelon form of the augmented matrix must show the 
  signal equation.

3.1-18  x+5x+6y=3, 5x+2y-10z=1, 8x+17y+8z=5. Find the general 
======  solution or else explain why there is no solution.

 The book reports "No Solution" which is correct. Display a frame
 sequence with documented toolkit steps swap, combo, mult, ending in
 a last frame that contains a signal equation "0=1" followed by the
 statement "No Solution" or "Inconsistent."

 MAPLE ANSWER CHECK.
  The reduced row echelon form of the augmented matrix must show the 
  signal equation.

3.1-26  y''-121y=0, y(0)=44, y'(0)=22, y(x)=A exp(11x)+Bexp(-11x)
======

 The constants are A=23, B=21 in the formula y=A exp(11x)+B
 exp(-11x). Follow 3.1 Example 8.

 It is not needed to know how to solve the differential equation, no
 is the differential equation used at all in the solution. This is
 because the general solution y of the differential equation has
 been supplied.

===end of 3.1 problem notes===