Math 54(R) Homework #4 Answer Key
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 =============================  Hill section 1.4  =============================

(#37 is at the end, because the solution is more involved.)

45.
     Simply multiply the two matrices together. If your answer simplifies to
     I (the identity matrix), you know that they are inverses.
     (Technically you have to multiply them in both orders: AB = I = BA.
     In fact, if AB = I, it follows that BA = I.)
     The following trig identities help you simplify:
     sin(-theta) = -sin(theta) [sin is odd ]
     cos(-theta) = +cos(theta) [cos is even]
     cos^2(theta)+sin^2(theta)=1.

46.
     Run through Gaussian elimination on this matrix.
     The first step depends on whether a=0 or not, so consider two cases.

     Case I: a != 0.
         This means that a is the first pivot, and elimination starts by
         subtracting some multiple (call it m) of the first row from the
         second row:
         [ a     b     ]
         [ c-m*a d-m*b ]
         The coefficient m is chosen so c-m*a=0, so one would use m=c/a.
         i.e., the resulting matrix is
         [ a b      ]
         [ 0 d-bc/a ]
         But d-bc/a = 0 because we're given ad-bc=0. So the bottom row is
         all zeros, meaning that this matrix will not be invertible.

     Case II: a = 0, but c != 0.
         In this case, the first pivot is c in the second row.
         So elimination begins with a row swap, yielding:
         [ a' b' ]   [ c d ]
         [ c' d' ] = [ a b ]
         You can continue with the elimination process from here.
         Or, you can observe that a'd'-b'c' = cb-da = -(ad-bc) = 0.
         Thus, after one step of elimination we're in case I, in which
         elimination is known to fail.

     Case III: a = 0 and c = 0
         In this case, the entire left column is zero, so the matrix is
         clearly singular.

37.
     (a)
         Easiest method: just plug into the 2x2 matrix inverse formula.
         This method doesn't generalize to bigger matrices, or part (b).

         A more general but more difficult method is to carry out the usual
         method for computing the inverse of a square matrix, with variables
         taking the place of numbers. Start by adjoining an identity matrix:
         [ a 0 | 1 0 ]
         [ c b | 0 1 ]
         We're given that a is nonzero, so the "a" is the first pivot.
         We subtract a multipe of the first row (say, m * 1st row) from the
         second to eliminate c:
         [    a  0 |  1 0 ]
         [ c-ma  b | -m 1 ]   chosen so that c-ma=0.
         In other words, we subtracted c/a times the first row from the second:
         [ a 0 |  1   0 ]
         [ 0 b | -c/a 1 ]
         Then divide the first row by a and the second row by b to get:
         [ 1 0 |  1/a  0   ]
         [ 0 1 | -c/ab 1/b ]

     (b)
         Method #1: Guess based on part (a).
         The formula has to be modified somewhat, because dividing a matrix
         by another matrix doesn't make any sense. The matrix version of
         "x/y" is either X Y^-1 or Y^-1 X, and it matters which one you
         choose!

         Fortunately, only the bottom-left entry is in doubt: 1/a should
         correspond to A^-1 and 1/b should correspond to B^-1.
         Unfortunately, the bottom left entry could be CA^-1B^-1, A^-1B^-1C...
         A pure guess-and-check method would force you to try 6 different
         arrangements, and once you found the right choice it would be a
         complete mystery why it worked.
         To improve your odds, just call the bottom-left entry X and solve
         for it!
         
         What we want is:
         [ A^-1 0    ] [ A 0 ]   [ I 0 ]
         [ X    B^-1 ] [ C B ] = [ 0 I ] (big identity matrix)
         But what we have is:
         [ A^-1 0    ] [ A 0 ]   [ A^-1A+0C A^-10+0B ]   [ I         0 ]
         [ X    B^-1 ] [ C B ] = [ XA+B^-1C X0+B^-1B ] = [ XA+B^-1C  I ]
         So what we really need is XA+B^-1C=0.
         =>  XA = -B^-1 C
         =>  X  = -B^-1 C A^-1

         Method #2: Generalize the process of Gaussian elimination.
         You can use the elimination method of part (a) here, as long as you
         observe one crucial rule: when you multiply a row by a factor M, the
         M has to go on the left. Why the left? Because that's where the
         matrices which represent row-operations have to go.
         The calculation would look like this:
         [ A 0 | I 0 ]
         [ C B | 0 I ]
         Subtract a multipe of the first row (say, M * 1st row) from the
         second to eliminate C:
         [    A  0 |  I 0 ]
         [ C-MA  B | -M I ]   chosen so that C-MA=0.
          ==>  MA = C  ==>  M = C A^-1
         [ A 0 |  I     0 ]
         [ 0 B | -CA^-1 I ]
         Multiply the first row by A^-1, second by B^-1 (on the left):
         [ I 0 |  A^-1       0    ]
         [ 0 I | -B^-1CA^-1  B^-1 ]