> with(LinearAlgebra):
> read("/afs/eos.ncsu.edu/users/k/kaltofen/www/courses/LinAlgebra/Maple/
> RefPkg/InitPkg.mpl");
> with(RefPkg);

  libname := /afs/eos.ncsu.edu/users/k/kaltofen/www/courses/LinAlg\

        ebra/Maple, "/usr/local/maple/lib"


       [E_I, E_II, E_III, MyDet, MyInverse, MySolve, Ref, XRef]

# 1. a
> M := <<x-1,2,0,-3>|<0,x,3,0>|<-3,2,x+2,0>|<2,0,0,x+1>>;

                      [x - 1    0     -3        2  ]
                      [                            ]
                      [  2      x      2        0  ]
                 M := [                            ]
                      [  0      3    x + 2      0  ]
                      [                            ]
                      [ -3      0      0      x + 1]

# 1. b I am going across the third row.
> M1 := <<x-1,2,-3>|<-3,2,0>|<2,0,x+1>>;
> M2 := <<x-1,2,-3>|<0,x,0>|<2,0,x+1>>;

                           [x - 1    -3      2  ]
                           [                    ]
                     M1 := [  2       2      0  ]
                           [                    ]
                           [ -3       0    x + 1]


                           [x - 1    0      2  ]
                           [                   ]
                     M2 := [  2      x      0  ]
                           [                   ]
                           [ -3      0    x + 1]

# Det(M) = -3*Det(M1) + (x+2)*Det(M2)
> -3*Determinant(M1) + (x+2)*Determinant(M2);
> simplify(%);

                    2                         3
                -6 x  - 48 - 18 x + (x + 2) (x  + 5 x)


                        2               4      3
                      -x  - 48 - 8 x + x  + 2 x

# 2. a
> A := <<1,0,2>|<-2,-1,4>|<0,3,1>>;
> Determinant(A);
> 

                              [1    -2    0]
                              [            ]
                         A := [0    -1    3]
                              [            ]
                              [2     4    1]


                                 -25

# So Det(A) is not 0, thus A is invertible.
# 2. b
> A1 := <<3,-2,4>|<-2,-1,4>|<0,3,1>>;
> A2 := <<1,0,2>|<3,-2,4>|<0,3,1>>;
> A3 := <<1,0,2>|<-2,-1,4>|<3,-2,4>>;

                              [ 3    -2    0]
                              [             ]
                        A1 := [-2    -1    3]
                              [             ]
                              [ 4     4    1]


                               [1     3    0]
                               [            ]
                         A2 := [0    -2    3]
                               [            ]
                               [2     4    1]


                              [1    -2     3]
                              [             ]
                        A3 := [0    -1    -2]
                              [             ]
                              [2     4     4]

# x1 = Det(A1)/Det(A), x2 = Det(A2)/Det(A), x3 = Det(A3)/Det(A)
# 2. c
> a1 := Determinant(A1);
> a2 := Determinant(A2);
> a3 := Determinant(A3);

                              a1 := -67


                               a2 := 4


                               a3 := 18

# x1 = 67 / 25, x2 = -4 / 25, x3 = -18 / 25
# 
# 3. This is not a vector space because it violates Axiom 6,
# (a+b)*v=a*v+b*v.  Here is an example.
# Let v be the vector [1,1]. Then 2*v=[2,1/2]. Since 2=1+1 and 
# 1*v+1*v=[1,1]+[1,1=[2,2] then 
# [2,1/2] = 2*v != 1*v+1*v=[2,2]. Thus Axiom 6 is violated.
# 
# 4.  To prove this is a subspace then I must show it is closed under
# addition and scalar multiplication.
# Addition: Let B,C be in NR(A). So A*B=0 and A*C=0.
# A*(B+C)=A*B+A*C=0+0=0. Thus 
# B+C is in NR(A).
# Scalar Multiplication: Let B be in NR(A) and r be in R. A*B=0 so
# A*(r*B)=r*(A*B)=r*0=0 and so
# r*B is in NR(A).
# So NR(A) is closed under addition and scalar multiplication and so it
# is a subspace of the n x n matrices.
# 
# Bonus. NR(A) is not equal to NL(A) for any n x n matrix A.
#  To prove this I only need to find a matrix A and a matrix B suct that
# B is in NR(A) and not in NL(A).
> A:=<<1,0>|<0,0>>;
> B:=<<0,1>|<0,0>>;

                                 [1    0]
                            A := [      ]
                                 [0    0]


                                 [0    0]
                            B := [      ]
                                 [1    0]

> A.B;
> B.A;
> 

                               [0    0]
                               [      ]
                               [0    0]


                               [0    0]
                               [      ]
                               [1    0]

> 
# So B is in NR(A) but B is not in NL(A) and thus NR(A) is not equal to
# NL(A) for any n x n matrix A.