MA 305 Spring 2K Homework 3 Solutions
> read("/afs/eos.ncsu.edu/users/k/kaltofen/www/courses/LinAlgebra/Maple/initlib.mpl"):
> with(refpkg):
> with(linalg):
Problem 1
(a.)
We are looking for a matrix A; and vector b; such that with the vector x = [x[1], x[2], x[3]]^T;, the linear system described by problem 1 can be written as A*x = b;.  The following matrix and vector do just that.
> A := matrix(3,3,[2,-1,1,4,-1,4,-2,1,2]);

                              [ 2    -1    1]
                              [             ]
                         A := [ 4    -1    4]
                              [             ]
                              [-2     1    2]

> b := vector(3,[-1,-2,-2]);

                          b := [-1, -2, -2]

(b.)
We know we can use a series of elementary matrices to reduce A; to row echelon form: E[k]*`...`*E[2]*E[1]*A = U;.  Then the transformation matrix T; is T = E[k]*`...`*E[2]*E[1];.  The following two matrices are the U; and T;, respectively, such that T*A = U; and U; is in row echelon form.
> U:=xref(A,3,'T','L');

                              [2    -1    1]
                              [            ]
                         U := [0     1    2]
                              [            ]
                              [0     0    3]

> print(T);

                            [ 1    0    0]
                            [            ]
                            [-2    1    0]
                            [            ]
                            [ 1    0    1]

Problem 2
Because T; is a product of elementary matrices, and all elementary matrices are invertible, T; must be invertible, and A = T^(-1)*U; or A = L*U; where L; is the inverse of T;.  The following two matrices are the corresponding L; and U;.
> print(L);

                            [ 1    0    0]
                            [            ]
                            [ 2    1    0]
                            [            ]
                            [-1    0    1]

> print(U);

                            [2    -1    1]
                            [            ]
                            [0     1    2]
                            [            ]
                            [0     0    3]

Problem 3
Now, from the original matrix system A*x = b; and the L*U; factorization of A;, we have L*U*x = b; or the two equations L*y = b; and U*x = y; which are easier to solve than the original system.  Using these two equations we find the following solution to the system. 
> y:=linsolve(L,b);

                           y := [-1, 0, -3]

> x:=linsolve(U,y);

                           x := [1, 2, -1]

Thus the solution to this system is x = [1, 2, -1]^T;.
Problem 4
Cramer's rule give us the solution to the same system using a different method.  Here we use the matrix A; and three other matrices A[1];, A[2];, andA[3]; such that A[j]; is created by replacing the j;-th column of A; by the column vector b;.  Then the i;-th component of the solution is x[i] = det(A[i])/det(A);.
> A1 := matrix(3,3,[-1,-1,1,-2,-1,4,-2,1,2]);

                              [-1    -1    1]
                              [             ]
                        A1 := [-2    -1    4]
                              [             ]
                              [-2     1    2]

> A2 := matrix(3,3,[2,-1,1,4,-2,4,-2,-2,2]);

                              [ 2    -1    1]
                              [             ]
                        A2 := [ 4    -2    4]
                              [             ]
                              [-2    -2    2]

> A3 := matrix(3,3,[2,-1,-1,4,-1,-2,-2,1,-2]);

                              [ 2    -1    -1]
                              [              ]
                        A3 := [ 4    -1    -2]
                              [              ]
                              [-2     1    -2]

> x[1] := det(A1)/det(A);

                              x[1] := 1

> x[2] := det(A2)/det(A);

                              x[2] := 2

> x[3] := det(A3)/det(A);

                              x[3] := -1

> print(x);

                              [1, 2, -1]

Thus the solution to this system is x = [1, 2, -1]^T;.