As usual Maple from Start Menu/Program Files/ or using the icon
in f:\math115 and load all the linear algebra progams.

Create a random 4x4 matrix B using this command:
> B:=RandomMatrix(4,4,generator=-7..6);
We will now set one element of B to be an unknown:
> B[3,2]:=x:
We can now see the new B using
> evalm(B);
Find Determinant(B) and use these commands to check the answer which makes 
the new B singular:
> q:=solve(Determinant(B)=0,x);
> B1:=subs(x=q,B);
> Determinant(B1);
Note: do not set "x:=q;" as then x will always have this value. 
Use x:='x'; to restore it as a variable.

Now set B[4,4]:=y; and repeat this to find the determinant of B in terms
of x and y.
Note what polynomial is in the denominator of your new q and deduce the 
value of y which will make your new B be singular whatever x is.
Check your answer using "subs".

We can calculate the adjoint of B using 
> Ba:=Adjoint(B);
Multiply B by Ba to note the appearance of the determinant and zeros.
What is the adjoint if x and y are values that makes B singular? 
What is the product of B and the adjoint in this case?

Now create a 3 by 3 matrix by evaluating: 
> A:=<<9, 6, -4>|<-9, -6, 6>|<-3, -3, 5 >>;

Use Maple to check the determinant of A and then form the matrix from which
we will get the eigenvalues:

> A1:=A-lambda*IdentityMatrix(3);

Use determinant row and column operations to create a situation from which
you could do a Laplace expansion and just have one 2x2 SubMatrix. Evaluate
the determinant of this submatrix and check it is the same when multiplied
in your laplace expansion as 
> pA:=Determinant(A1); 
which is the so-called characteristic polynomial of A.
 
To manipulate polynomials you can use factor, expand, sort, etc. 
> factor(pA);
> expand((3-lambda)*Determinant(A3));
 Note that the determinant of A appears as the unit coefficient in f.
You can also plot pA to see the locations of the eigenvalues using

> plot(pA,lambda=1.5..4);

Find the eigenvectors belonging to the eigenvalue 2 using 

> v1:=LinearSolve(subs(lambda=2,A1),<0,0,0>);

and verify that all vectors satisfy Av = lambda v as expected.
Find both eigenvectors of the eigenvalue 3 similarly.
What are the solutions if you try to substitute a non eigenvalue?

Try to repeat the above procedure for 
> C:= <<-6, 2, -3>|<-13, 15, -15>|<-2, 8, -7>>;
> C1:=C-lambda*IdentityMatrix(3);
but note that each way you try to do operations on C1 no cancellation occurs.

Use
> Eigenvalues(C);
to note what kind of factors you should get and try to create one. Once you
have failed to create any zeros, just evaluate 

> db:=factor(Determinant(C1)); 

Find one of the three eigenvectors by RowOperation and check for it in

> (v,P):=Eigenvectors(C);

Extract another eigenvector from P and check it has the expected eigenvalue.

Make the diagonal matrix of eigenvalues using

> D1:=DiagonalMatrix(v);

Check that the product and P and D1 is the same as that of C and P.