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

The matrix underlying the recurrence b(n+1)= 3*b(n) + 28*b(n-1) is
> B:=<<3,1>|<28,0>>;

Diagonalise B utilising
> (v,P):=Eigenvectors(B);
and notice that its eigenvectors have the special "powers
of eigenvalues" formula they are supposed to have.

Use v and P to form the nth power of B.

Verify that 
> MatrixPower(B,n); 
gives an equivalent answer.

Now suppose that the first two values in the sequence are b(0)=1 and b(1)=2 .
Calculate b(2), b(3) and b(4) using the recurrence then use the power matrix to
get b(n+1) and b(n) and check with your values with n=3.

Find initial values for the sequence which would give a solution with fewer 
than 2 eigenvalue powers in the final solution.

Create the 3 by 3 matrix A corresponding to the recurrence 
a(n+1)=-5*a(n)+2*a(n-1)+24*a(n-2)
when
a(0)=5,a(1)=-30,a(2)=120

Find the eigenvalues by solving when this matrix is singular:
> A1:=A-lambda*IdentityMatrix(3);
Try a few row/col operations on A1 and see how difficult it is to get nice 
factors which will cancel, and then just do a laplace expansion of A1 and 
see how the coefficients from the recurrence re-occur in the determinant.
Check your answer with
> f:=sort(Determinant(A1));

Factor f using the Maple "factor" command and then create the
predicted eigenvectors and verify that they all satisfy Av = lambda v .
Thus diagonalise A and get the nth power of A as before and multiply by
the inital values to get the final answer. Check that a(4) is the same
by both methods of finding it.

Quickly find the eigenvalues and eigenvectors of this matrix 
> E:=<<-5, -2, 4>|<48, 5, -24>|<4, -4, 1>>;
and then use MatrixPower to check the general form of the nth power contains
the eigenvectors as coefficients of the powers of the eigenvalues and see
when you multiply the power of the matrix by any vector on the right these
are preserved.

You can plot points using this command:
> plot([[-1,6],[2,3],[3,7]],style=point, symbolsize=30,colour=blue);
Set up the 3x3 matrix and solve the system of equations to get the 
coefficients of a quadratic polynomial which you can then plot using:
> plot(a*x^2+b*x+c,x=-2..4);
Note that the curve passes through all three points.

Repeat with polynomials of degrees 1 and 3 and see what happens.

You can plot the family of polynomials of degree 3 like this:

> with(plots);
> p3j:=((2*d-5)*x^3 + (35-8*d)*x^2+(2*d-32)*x + 12*d)/12;
> animate( plot, [p3j,x=-2..4], d=-10..10, trace=5, frames=50 );

To animate the plot you click on the image and then press the 
"play" button that is now in the toolbar at the top. Note how every 
cubic curve passes through all 3 specified points.