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

The matrix underlying the recurrence c(n+1)= 6*c(n) + 7*c(n-1) - 60*c(n-2) is
> C:=<<6,1,0>|<7,0,1>|<-60,0,0>>;

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

Form Ck as the kth power of C and now suppose that the first three values in the 
sequence are c(0)=8, c(1)=-85 and c(2)=27. 
Multiply Ck by the initial values vector, remembering to use the correct order.

Calculate c(3) and c(4) using the recurrence then use the power matrix to
get c(k) and check with your values with k=3 and 4.
At which point is c(k) below zero for the last time?

Find initial values for the sequence which would give a solution with less 
than three different eigenvalue powers in the final solution for c(k).


You can plot points using this command:
> plot([[-1,5],[2,4],[3,-1]],style=point, symbolsize=30,colour=blue);
The matrix for a quadratic underlying this system of equations is
> E2:=<<1,4,9>|<-1,2,3>|<1,1,1>|<5,4,-1>>;

Find the solutions to this system of equations to get the 
coefficients of a quadratic polynomial which you can then plot using:
> plot(your polynomial here,x=-1..3);
Note that the curve passes through all three points.


Repeat with polynomials of degree 3 through these points:
    [-1,3], [-2,6], [4,-12]
the system of equations in this cases is:
> E3:=<<-1,-8,64>|<1,4,16>|<-1,-2,4>|<1,1,1>|<3,6,-12>>;
Solve for a*x^3+b*x^2 + c*x +d using
> LinearSolve(E3,method='none',free=t);
to see that a=-t, b=t, c=10*t -3, d=8t.

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

> with(plots);
> p3j:=(-t*x^3 + t*x^2+ (10*t-3)*x + 8*t);
> animate( plot, [p3j,x=-3..5], t=-5..5, trace=4, 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.When t=0 the
cubic curve is actually a straight line.

Pick 5 random points and create the system of equations for 
a straight line through them as follows:
> xv:=RandomVector(5,generator=-4..4);
> F:=< xv | <1,1,1,1,1>>;
> yv:=RandomVector(5,generator=0..7);

Use this to plot your points:
> plot((xv,yv),style=point, symbolsize=30,colour=red);

We can now create the best fit straight line as follows:
> FTF:=Multiply(Transpose(F),F);
> FTy:=Multiply(Transpose(F),yv);
Plot the solution to FTF times v = FTy, which is the best fit line,
by using RREF, inverses or by doing 
> LinearSolve(FTF,FTy); 

Now repeat with the same xv and yv to plot the best fit quadratic 
to your data.

Create a 4x4 matrix A with an eigenvalue of multiplicity 3 as
described in class and ensure it has only two eigenvectors of 
the eigenvalue of multiplicity 3.
Now create a random invertible 4x4 matrix Q and verify that this 
matrix B has the same eigenvalues and multiplicities, but more 
complicated eigenvectors than A:
> B:=Multiply(Multiply(Q,A),MatrixInverse(Q));