As usual Maple from Start Menu/Program Files/ or using the icon
in f:\math 1204 and load all the linear algebra progams.
Please randomise to start with:
> Seed:=randomize():

Create a random 3x3 matrix A using this command:
> A:=RandomMatrix(3,3,generator=2..11);

We can evaluate the determinant directly using
> Determinant(A);

We can extract the cofactors using the command SubMatrix;
For instance
> C12:=SubMatrix(A,[2,3],[1,3]);
chooses to start with matrix A and keep the rows 2 and 3 
and the columns 1 and 3. Note that order is important.

Extract three submatrices from one row like this and evaluate the
sum of the products of their determinants with the +- matrix and 
the corresponding entries of A to give the Laplace expansion.

Repeat with three submatrices from a column and verify that
they again give the determinant.

Use RowOperation (and ColumnOperation which works the same)
to produce as many zeroes as possible from A, and note how
the determinant appears this time.

Use the determinants of the submatrices of A to form what
you believe will be the adjoint of A and call it J.
See if you get what you should when you Multiply A and J.
Maple can produce what J should be using
> Adjoint(A);

Form the 3x6 matrix with A with the identity matrix adjoined;
> N:=<A | IdentityMatrix(3) >;
Use RowOperation and Pivot to create the inverse of A.

Check that 
> Adjoint(A)/Determinant(A);
is the same as the inverse of A (if it exists)

Create a 3x3 matrix E with rank less than 3 and verify what
> Adjoint(E);
and 
> Determinant(E);
are and see what happens when you ask for 
> MatrixInverse(E);

Create a random 4x4 matrix B using this command:
> B:=RandomMatrix(4,4,generator=-7..6);
We will now set three elements of B to be unknown:
> B[3,2]:=x: B[1,3]:=x-2: B[4,1]:=2*x+3:
We can now see the new B using
> evalm(B);

Find Determinant(B) as a polynomial in x and use these commands 
to check an answer which makes the new B singular:
> q:=evalf(solve(Determinant(B)=0,x));
Note that you may have some complex numbers as solutions.
Assuming you have a real solution in position 1 of q, use:

> B1:=subs(x=q[1],B);
> Determinant(B1);
Is this the value you expected?

Note: do not set "x:=q;" as then x will always have this value. 
Use x:='x'; to restore it as a variable.