As usual run Maple and load all the linear algebra progams.

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);

ColumnOperation works in just the same way as RowOperation does.
Use a pair of determinant row and column operations to create two zeros such that you could do a Laplace expansion with two zeros as coefficients. 

Repeat with a different row/column pair to identify a different factor.

We can find the polynomial whose roots are 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.

To manipulate polynomials you can use factor, expand, sort, etc. 
> factor(pA);
Verify that both of the different linear polynomials in lambda are included.

You can also plot pA to see the locations of the eigenvalues using
> plot(pA,lambda=1.5..4);

We can also get the system of equations to solve using this matrix:
> Aj:=<subs(lambda=2,A1)|<0,0,0>>;
Pivot this matrix to create a row of zeros and read off the solution.

Compare what you just got with this vector solution:
> v1:=LinearSolve(subs(lambda=2,A1),<0,0,0>,free=s);

and verify that v1 when multiplied by A on the left gives the expected result (for any s).

Find two different eigenvectors of the eigenvalue lambda=3 similarly and check them.
3 is an eigenvalue of multiplicity 2 since it has two different vector solutions.

What are the homogeneous solutions if you try to substitute a non eigenvalue into A1?


Try to repeat the row and column operations procedure for 
> B:=<<32, -36, -18>|<27, -31, -18>|<-6, 6, 2 >>; 
> B1:=B-lambda*IdentityMatrix(3);

Some row/column operations give two easy zeros but some don't. Find one which works.

Now try with this matrix
> 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. If after 5 minutes if you have failed to create two zeros in a row, 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/or Pivot 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.

Create a matrix which has the eigenvalues from matrix w as follows:
> randomise():w:=RandomVector(3,generator=rand(1..9));
> Dw:=DiagonalMatrix(w);
> Q:=RandomMatrix(3,3,generator=rand(-3..4));

We choose Q as a random matrix for the eigenvectors and (if Q is 
non-singular) use diagonalisation to make J from Dw as follows:
> J:=Multiply(Multiply(Q,Dw),MatrixInverse(Q));

Check that the eigenvalues and eigenvectors of J are as you expect.

Maple can create the kth power of a matrix automatically:
> Jk:=MatrixPower(J,k);

> Dk:=MatrixPower(Dw,k);
Check that when you use k=2 you get the same as
> Multiply(Dw,Dw);

Multiply P and a diagonal matrix of the powers of the eigenvalues, and 
the inverse of P to get Bk, the kth power of B.

> Bk:=Multiply(Multiply(Q,MatrixPower(DiagonalMatrix(w),k)),MatrixInverse(Q));

This should be something like Jk, but might be a bit different in some ways. 
You can see they are the same using 
> simplify(Jk-Bk);

Check that Jk and Bk are the same for small values of k:

> subs(k=2,Bk); subs(k=2,Jk); Multiply(J,J);

Check that this holds also for k=-1 and 0, giving the inverse and the identity.