4246.sagews

M=MatrixSpace(QQ,3)

I=M([1,0,0,0,1,0,0,0,1]); I

[1 0 0] [0 1 0] [0 0 1]

MC=MatrixSpace(QQ,3,1)
b=MC([0,1,0]);b

[0] [1] [0]

# Working on problem 14a in Rorres section 1.6

A=M([1,3,5,-1,2,0,2,5,4])
A

[ 1 3 5] [-1 2 0] [ 2 5 4]

C=A\b; C

[-13/25] [ 6/25] [ -1/25]

C1=A.solve_right(b)
C1

[-13/25] [ 6/25] [ -1/25]

C2=A.augment(b)
C2

[ 1 3 5 0] [-1 2 0 1] [ 2 5 4 0]

C2.echelon_form()

[ 1 0 0 -13/25] [ 0 1 0 6/25] [ 0 0 1 -1/25]

A_inv=A.inverse()
A_inv

[ -8/25 -13/25 2/5] [ -4/25 6/25 1/5] [ 9/25 -1/25 -1/5]

C3=A_inv*b
C3

[-13/25] [ 6/25] [ -1/25]

C4=A.augment(I)
C4

[ 1 3 5 1 0 0] [-1 2 0 0 1 0] [ 2 5 4 0 0 1]

C4.echelon_form()

[ 1 0 0 -8/25 -13/25 2/5] [ 0 1 0 -4/25 6/25 1/5] [ 0 0 1 9/25 -1/25 -1/5]

#Notice that the result above is I | A_inv
#Notice that we can get the solution to Ax = b as C, or C1, or the rightmost column of C2.echelon_form() or C3