Start Maple from Start Menu/Program Files or using the icon in f:\math115.

Load the linear algebra progams with the command:
>    with(linalg):

Find the rank of this matrix, and check that the nullspace, column space and row
space have the expected dimension using row operations and/or pivoting.

> F := matrix([[-3, 4, -6, 4], [1, -8, -1, -10], [4, -12, 5, -14], [-2, -4, -7, -6]]);
> Fa:=augment(F,matrix(4,1,[a,b,c,d]));

Check your answers using the commands nullspace, rowspace and colspace, by expressing 
your vectors in terms of the given ones and vice versa.
Work out the equation for the rows in terms of the 3rd and 4th row and check your answer. 
Repeat for the first and third columns. You can use 

> Fc2:=transpose(matrix([c1,c3,c2]));

to form the augmented matrix if your columns are to get c2 in terms of c1 and c3 .

Choose a vector which you think is not in the rowspace and prove that it is independent
to your two basis row vectors. Choose one more vector which is hopefully independent to 
this new set of 3 independent vectors and check that the determinant of the matrix made 
up of these 4 vectors is non zero but that the determinant of your three independent 
vectors and any of the other rows of F is zero.


Find the three spaces again for this matrix:
 
> G:=matrix([[2, 1, 2, -2], [-2, 1, -1, 0], [2, -1, 2, 0], [2, -1, -2, 0], [0, 1, -2, -1]]);

Verify the relationship between the spaces is the same again, so that the numbers in
the row and col spaces come from the reduced nullspace matrix, and the dimensions
of the row and col spaces are the same as the rank again.

Create independent vectors as before, but this time show that we can get a set of 5 
independent vectors for the colspace, but only 4 for the rowspace. 

Prove that the space spanned by the vectors
 [-2, 2, -4] and [4, 6, -2] is the same as that spanned by [-3, -6, 3] and [-1, -3, 2].
(transpose(stackmatrix(v1,v2,v4)); may prove useful here)

Find [41, 72, -31] in terms of both bases, and find the vector orthogonal to this space
and show it is not in the space, and hence independent.