How do you solve the system #3y + 2z = 4#, #2x − y − 3z = 3#, #2x + 2y − z = 7#?

1 Answer
Nov 27, 2015

Answer:

# [(x), (y), (z)] = t [(-7), (4), (-6)] + [(9/2), (0), (2)] , forall t in mathbb{R}#

Explanation:

#[(0, 3, 2), (2, -1, -3), (2, 2, -1)] * [(x), (y), (z)] = [(4), (3), (7)]#

#L_3' := L_2 - L_3 : [2-2, -1-2, -3 +1] * [(x), (y), (z)] = 3 - 7#

#L_3' : [0, -3, -2] * [(x), (y), (z)] = -4#

#L_3'' = L_1 + L_4: [0, 3-3, 2-2] * [(x), (y), (z)] = 4 - 4#

This is true for all #(x, y, z) in mathbb{R}^3#

#L_1 Rightarrow z(y) = (4 - 3y)/2#

#L_2 Rightarrow 2x - y - 3 (4 - 3y)/2 = 3# and we want x(y).

#4x - 2y - 12 + 9y = 6#

#4x = 18 - 7y#

# [(x), (y), (z)] = [((18-7y)/4), (y), ((4 - 3y)/2)] = y [(-7/4), (1), (-3/2)] + [(18/4), (0), (4/2)] #

#y = 4t#