Start Maple from the desktop, Start Menu/Program Files/ 
or using the icon in f:\math 1204

Load all of the linear algebra progams with the command:
>    with(LinearAlgebra);

Define
> J:=Matrix([[-3, 1, -4], [2, 4, 3], [2, -1, 0]]);
> w:=Matrix(3,1,[1,8,4]);

We can augment a column to our matrix J (but not make a dotted 
line unfortunately) using the "|" symbol:
> J0:=< J | w >;

Use the command Pivot (use help to find out how it works)
to produce two zeros in column 2, then continue using Pivot 
until you have six zeros on the left of the dotted line.

Try to see what the answer to this system of equation was as
> v:= < number, number, number >;
and check that you get w with the command "Multiply" as follows:

> Multiply(J,v);

If it isn't right, try predicting again until it works.


We can also use the same pivots to produce the inverse:

Create 
> JI:=< J | IdentityMatrix(3) >;

and use the same pivots and then Row Operations to produce
the Identity Matrix on the left where J was.

You should get the same matrix as from
> Ji:=MatrixInverse(J);

Check that 
> Multiply(Ji,w);
gives v as expected.

Repeat the above steps to try to find the inverse of a 4x4.
> L:=RandomMatrix(4,4,generator=rand(-1..1));
First see if L has rank 4:
> Rank(L);
If it does, repeat until you get a new L that has rank < 4.

Pivot and use row operations on
> LI:=< L | IdentityMatrix(4) >;
until you can't go any further and see how it corresponds to
the rank of L.


Now we will look at how Maple copes with multiple solutions.
Create a random matrix K0 to represent a 3x4 matrix equation
and its solution vector:
> K0:=RandomMatrix(3,5,generator=rand(-4..4));

Use Pivot and RowOperation to produce as many zeros and ones
as possible in the first four columns. Unless you have a row
of zeros you will have just one column you couldn't pivot.

Now use your brain and/or pen and paper to predict the 
particular and homogeneous solutions and input them using
> h:= < number, number, number, number >;
and 
> p:= < number, number, number, number >;
with the "number"s being your found values.

Note that you can extra the 3x4 matrix K as follows
> K:=DeleteColumn(K0,5);

Start again with the same K0 but leave a different column 
unpivoted this time. Continue until you again have a different 
particular solution and notice that the homogeneous solution 
involves the same ratios between numbers.

Notice how similar answers appear with the commands:
> ReducedRowEchelonForm(K0);
and 
> LinearSolve(K0,free='t');

Now find two homogeneous solutions of 
> Q0:=RandomMatrix(2,5,generator=rand(-4..4));
and check them and the particular solution by multiplication.



Create a random 4x3 matrices M using 
> M:=RandomMatrix(4,3,generator=rand(5..13));
and then create N as a 3x4 matrix, similarly.

Multiply M and N, firstly with
> Multiply(M,N);
and then with
> Multiply(N,M);
and notice that these two are not the same, it is important
to remember that it matters with matrices which order they 
are being multiplied in.

Maple will refuse to multiply matrices of the wrong size,
try to multiply M by itself, N by J and J by N and see if 
it works when you expect it to. The purple response tries
to explain what went wrong.