How do you solve the system #x-2y+8z=-4#, #x-2y+6z=-2#, #2x-4y+19z=-11#?

1 Answer
Vinícius Ferraz
Dec 8, 2015

#(x, y, z) = t V + W, forall t in RR#

Explanation:

#((1, -2, 8), (1, -2, 6), (2, -4, 19)) cdot ((x), (y), (z)) = ((-4), (-2), (-11))#

#((1, -2, 8), (1-1, -2+2, 6-8), (2-2, -4+4, 19-16)) cdot ((x), (y), (z)) = ((-4), (-2+4), (-11+8))#

#((1, -2, 8), (0, 0, -2), (0, 0, 3)) cdot ((x), (y), (z)) = ((-4), (2), (-3))#

#z = -1# and #x - 2y - 8 = -4 Rightarrow x = 2y + 4#

#((x), (y), (z)) = ((2y + 4), (y), (-1)) = y cdot ((2), (1), (0)) + ((4), (0), (-1)) #

Related questions
See all questions in Gaussian Elimination
Impact of this question
1457 views around the world
You can reuse this answer
Creative Commons License