4038.sagews

# Section 1.5

# 9 (Find solutions to the homogeneous equation Ax=0)
A = matrix(QQ,[[3,-6,6,0],[-2,4,-2,0]])
A

[ 3 -6 6 0] [-2 4 -2 0]

A.rref()

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

# Therefore, 
# x_1 = 2x_2
# x_2 is free
# x_3 = 0
x_1, x_2, x_3 = var('x_1,x_2, x_3')

# In vector form, this is 
x = vector(SR,[2*x_2,x_2,0]); x

(2*x_2, x_2, 0)

# 17 Find solutions to Ax=b, and write in parametric form.
A = matrix(QQ,[[2,2,4,8],[-4,-4,-8,-16],[0,-3,-3,12]])
A

[ 2 2 4 8] [ -4 -4 -8 -16] [ 0 -3 -3 12]

A.rref()

[ 1 0 1 8] [ 0 1 1 -4] [ 0 0 0 0]

# x_3 is the free parameter. x_1 = 8-x_3, x_2 = -4 -x_3. In vector form we get
x = vector(SR, [x_1,x_2,x_3])
x

(x_1, x_2, x_3)

# Thus,
x = vector(SR, [8,-4,0]) + x_3 * vector(SR,[-1,-1,1])
x

(-x_3 + 8, -x_3 - 4, x_3)