Question

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

Answer 1

`[(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`