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

Create a matrix B and its eigenvalues and eigenvectors:
> B:=Matrix([[6, -2, 1], [7, -3, 1], [-4, 2, 1]]);
> (v,P):=Eigenvectors(B);

Make vk a vector of the kth powers of the eigenvalues (in the same order
as v) and make Dk a diagonal matrix of them, then multiply P and Dk and 
the inverse of P to get Bk, the kth power of B.
> Bk:=Multiply(Multiply(P,MatrixPower(DiagonalMatrix(v),k)),MatrixInverse(P));

Check that 
> subs(k=1,Bk);
and
> subs(k=0,Bk);
give the expected answers, and that
> subs(k=-1,Bk);
does too.

Check your answer for Bk with 
> MatrixPower(B,k);

Multiply B by itself to get its square and cube and check that
they match the matrices formed by the appropriate substitution of k.

Create a matrix which has the eigenvalues from w:
> randomise():w:=RandomVector(3,generator=rand(1..9));
> Dw:=DiagonalMatrix(w);
Choose a random invertible matrix for the eigenvectors Q and use 
diagonalisation to make J from Dw.

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

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

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 only 
two 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 3x3 matrix 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);
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=-2..4], t=-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.

Show that there is a solution for a straight line through the points:
> E1:=<<-1,-2,4>|<1,1,1>|<3,6,-12>>;

Pick 5 random points and create the system of equations for 
a straight line through them:
> 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((x,y),style=point, symbolsize=30,colour=red);

And 
> FTF:=Multiply(Transpose(F),F);
> FTy:=Multiply(Transpose(F),y);
Plot the solution to FTF times v = FTy, which is the best fit line,
and repeat to plot the best fit quadratic.