-- Solutions to Matlab #1

 
>> rand('state',sum(100*clock));


-- Problem 1

>> A = rand(5,7);
>> rref(A)

ans =

    1.0000         0         0         0         0    0.5495   -1.5588
         0    1.0000         0         0         0    0.1922   -0.0387
         0         0    1.0000         0         0   -0.2642    0.8816
         0         0         0    1.0000         0    0.1217    2.3727
         0         0         0         0    1.0000    0.0616   -0.0866

-- Since there are columns (6 and 7) without pivots the columns of A are linearly dependent.
--  This also follows from Theorem 8 on page 69 since there are more columns than rows.

-- By Theorem 4 page 43 the Span of the columns of A is all of R^5 since there is a pivot in each row.



-- Problem 2

>> u = rand(7,1)

u =

    0.3968
    0.0669
    0.6684
    0.9432
    0.4518
    0.2227
    0.1300

>> b = A*u;
>> x = A\b

x =

    1.1566
    0.1255
    0.2491
         0
    0.5010
         0
    0.5389

--  Note that x is not equal to u, yet both are solutions to Ax=b.  
--  This happened because there is more than one solution to Ax=b,
--      since there are free variables
-- To get all solutions put the augmented matrix in reduced echelon form.

>> rref([A b])

ans =

  Columns 1 through 7 

    1.0000         0         0         0         0    0.5495   -1.5588
         0    1.0000         0         0         0    0.1922   -0.0387
         0         0    1.0000         0         0   -0.2642    0.8816
         0         0         0    1.0000         0    0.1217    2.3727
         0         0         0         0    1.0000    0.0616   -0.0866

  Column 8 

    0.3165
    0.1047
    0.7242
    1.2788
    0.4543

-- From this we can read off the solution:

--      x1 =  0.3165 -  0.5495 x6   + 1.5588 x7
--      x2 =  0.1047 -  0.1922 x6   + 0.0387 x7
--      x3 =  0.7242 + 0.2642 x6    - 0.8816 x7
--      x4 =  1.2788 -  0.1217 x6   -  2.3727 x7
--      x5 =  0.4543 -  0.0616 x6  +  0.0866 x7
--      x6 =  any scalar
--      x7 =  any scalar

-- By the way, here is an alternative.
-- The following command gives two linearly independent vectors
--    which span the set of solutions to the homogeneous equation Ax = 0.

>> N = null(A)

N =

   -0.4194    0.4484
   -0.1623   -0.0078
    0.1976   -0.2585
   -0.1770   -0.7840
   -0.0497    0.0216
    0.8509    0.1059
    0.0309    0.3250

--  Then the general solution is  x + s * N(:,1) + t * N(:,2) where s and t are arbitrary.


-- Problem 3

-- First set up the augmented matrix and then find its reduced echelon form.

>> C = [1+i, 2-i, 3i, 7-5i; 2, 0, 1-i, 4i; 1, 4i, 1+3i, -5+7i];

>> rref(C)

ans =

   1.0000                  0                  0             1.0000 + 1.0000i
        0             1.0000                  0             3.0000 + 1.0000i
        0                  0             1.0000            -2.0000   

-- From this we see there is just one solution  x1 = 1+i,  x2 = 3+i, x3 = -2.


-- Problem 4

-- Again take rref of the augmented matrix.     

>> rref([ 1+i, -3i, 1, 5-4i; 3-i, -6-i, 3, 1+2i])

ans =

   1.0000                  0             3.5000 - 0.5000i  15.5000 + 4.5000i
        0             1.0000             1.0000 - 1.0000i   8.0000 - 2.0000i

--  In this case we have a free variable x3 and there will be infinitely many solutions.
--  We can read off the solutions:

--  x1 = 15.5 + 4.5i - (3.5-.5i) x3
--  x2 =    8 - 2i  - (1-i) x3
--  x3 = any complex number.