As usual Maple from Start Menu/Program Files/ or using the icon
in f:\math115 and load all the linear algebra progams.
Please randomise again:
> Seed:=randomize():

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

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 with two zeros as coefficients. 
ColumnOperation works in exactly the same way as RowOperation.

We can find the polynomial which contains the eigenvalues using:
> pA:=Determinant(A1); 
which is the so-called characteristic polynomial of A.
Note that the coefficient of pA without any lambda in is the determinant of A.


Extract the 2x2 SubMatrix as S and evaluate
the determinant of S and check it is the same when multiplied
in your laplace expansion. If the non-zero entry of the row or column was (k-lambda), use
> expand((k-lambda)*Determinant(S));
and see that you get the same polynomial as pA (or the negative of it).

To manipulate polynomials you can use factor, expand, sort, etc. 
> factor(pA);


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 and check.
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 easy cancellation occurs.

Use
> Eigenvalues(C);
to note what kind of factors you should get and try to create one. After 5 minutes
if you have failed to create any zeros, just evaluate 
> db:=factor(Determinant(C1)); 
to reassure yourself the answers Maple gave are indeed correct.

Find one of the three eigenvectors in C using RowOperation and check for 
it or a multiple of it in the columns of P in

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

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

Make a diagonal matrix of eigenvalues using

> D1:=DiagonalMatrix(v);

Check that P and D1 multiplied is the same as C and P multiplied in that order.

Check these compound matrices have the expected values and vectors:
> Eigenvalues(k*C);
> Eigenvalues(MatrixInverse(C));
> Eigenvalues(Transpose(C));

Think about what these will always work in this way.