Math211.1.5

129 days ago by Jesse_Berwald

# 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]
[ 3 -6  6  0]
[-2  4 -2  0]
A.rref() 
        
[ 1 -2  0  0]
[ 0  0  1  0]
[ 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)
(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]
[  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]
[ 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)
(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)
(-x_3 + 8, -x_3 - 4, x_3)