As usual, start Maple and load the linear algebra progams with the command:
>    with(LinearAlgebra);

Create a random 4x4 matrix F with entries between -1 and 2 and make sure that Rank(F) = 4. 
We will now, given v:=<5,-1,3,2>; we will solve the matrix equation
       F<x1, x2, x3 ,x4> = v;

First solve it by augmenting the right hand matrix: 
>  F1:=< F | v );
Use ReducedRowEchelonForm(F1) to find the solution then
use a series of RowOperations to do the row reductions on F1 one by one 
to get to the same result.

Using the command MatrixInverse verify your result by multiplying the inverse and v.

Verify that Maple's inverse is correct by reusing your row operations on the 4x8 matrix made by augmenting I4:=IdentityMatrix(4); to F



Using
> C := Matrix(<<-2|-1|3|0|2|-2>, <-1|-1|-1|-2|2|0>, <0|-1|1|3|2|2>, <-1|3|-2|1|-1|0>, <-2|-1|3|0|2|-2>>);
and
> w:= <3,3,-6,5,3>;
Use RowOperation to find the row reduced matrix, verify that
> x:=LinearSolve(C, w, 'free=s');
gives a matrix that satisfies the equation
and check that <4,7,2,-3,5,-1> is in both of your solution sets.
(You can use the function subs and square brackets to index the s variables
to substitute values in.)


Form the 3x3 zeroes matrix and see how Maple reports it has no inverse.
>    Z:=Matrix(3,3,0);
Compare that message to the one when you try to find the inverse of C.

Generate several 2x2 matrices and find two whose inverses do exist.

Multiply these inverses and check that the relation for the
inverse of a product is the product of the inverses in the other order.
Check this relation still holds for three of your matrices multiplied.

Check that there is no easy relation between inverse(A+B) and their inverses

Find two 3x3 matrices C and W such that C and W have inverses but (C+W) doesn't
(note that we can't use D as a variable as it is used for differentiation)

Find another two inverses which do not have inverses but which when added 
make a matrix which does have an inverse.

Check that the inverse of the cube of F is the cube of the inverse of F.