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

We are now ready to do some matrix algebra...
Load all of the linear algebra progams with the command:
>    with(LinearAlgebra);

All the new commands listed are now known by Maple, we will use some.

We can create a 2x3 matrix by using the following command
>    A:=Matrix([[-5,-2,1],[3,1,-2]]);
It gives the matrix in terms of its rows.

Notice that we again use ":=" to assign a value to A. I recommend using
a capital letter for all matrices, but Maple will not insist on that. 
Recall that Maple is case sensitive, so A is different from a.

All of the LinearAlgebra commands also begin with a capital letter, 
although "matrix" is also defined, it will not always work properly.

Remember that to know more about a Maple command, type ? and the command e.g.:
>    ?Matrix
to exit from the help page, close the subwindow, or use the 
Window menu to move between them.

We can do the three different row operations as follows:
> A1:=RowOperation(A,[1,2]);
will swap row 1 and row 2 of A and make it into a new matrix A1.
> A2:=RowOperation(A1,2,-1);
would take this new A1 and multiply row 2 by -1 and make A2 
> A3:=RowOperation(A2,[2,1],-3);
will replace row 2 of A2 by itself minus 3 times row 1 to make A3

Note that this last operation will not give you the desired 0 in row 2
of column 2, so change this operation to make an A4 with a 0 there, then 
continue to reduce your matrix to have two zeros on the left hand side.

Next we create a random matrix B, with integer entries between -4 and +5

>    Seed:=randomize(): B:=RandomMatrix(2,3,generator=rand(-4..5));

The 2 and 3 are the numbers of rows and columns of the matrix, the -4 
is the lowest possible integer allowed and 5 is the highest possible.

Use RowOperation to produce as many zeros as possible from B and check
your final answer with
> ReducedRowEchelonForm(B);


Define
> K:=Matrix([[-1, 5, 2], [-4, 2, -4], [1, 1, 2]]);

We can augment a column to our matrix K (but not make a dotted 
line unfortunately) using the "|" symbol:
> K0:=< K | Matrix(3,1,[-1,-4,1]) >;

Use RowOperation to find the solution to this system of equations.

Check your answer with
> ReducedRowEchelonForm(K0);

Create, using Maple, a random 3x3 matrix J with integers between 1 and 9
and use RowOperation to get J0:=<J | Matrix(3,1,[3,-2,4])>; to RREF 


Now define 
> L:=Matrix([[2, 1, -3, -1], [1, 0, -4, 3], [0, -3, -3, -1]]);
> y:=Vector([8, 16, -6]);
and make
> L0:=<L | y>;
as the system of equations we are planning to solve.

Now we will use Pivot, which is just a set of row operations which uses 
one entry to make all others in that column zero.

Try
> L1:=Pivot(L0,1,4);
and find a set of Row Operations which give the same matrix.

Starting again with L0, choose appropriate entries to pivot around and create an 
equivalent of Reduced Row Echelon Form.
(Remember to use the RowOperation which multiplies rows by a number to make a 1)

Check your answer against Maple's:
> ReducedRowEchelonForm(L0);
Note the same ratios of numbers appear, even if the numbers are different.

Pivot by a different set of positions and see the same numbers appear again.