How do you solve using gaussian elimination or gauss-jordan elimination, $2 x + y - z = - 2$ $2x + y - z = -2$ , $x + 3 y + 2 z = 4$ $x + 3y + 2z = 4$ , $3 x + 3 y - 3 z = - 10$ $3x + 3y - 3z = -10$ ?

Question

2063 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-20T13:44:36

$(x, y, z) = (\frac{4}{3}, - \frac{4}{3}, \frac{10}{3})$ $(x,y,z)=(4/3,-4/3,10/3)$

So let's make the matrix of the system, which is :

$[(2,1,-1,-2),(1,3,2,4),(3,3,-3,-10)]$

Now let's do our calculations :

$[(2,1,-1,-2),(1,3,2,4),(3,3,-3,-10)]rarr[(1,-2,-3,-6),(1,3,2,4),(0,-6,-9,-22)]rarr$

$[(1,-2,-3,-6),(0,5,5,10),(0,-6,-9,-22)]rarr[(1,-2,-3,-6),(0,1,1,2),(0,-6,-9,-22)]rarr$

$[(1,-2,-3,-6),(0,1,1,2),(0,0,-3,-10)]rarr[(1,0,-1,-2),(0,1,1,2),(0,0,1,10/3)]rarr$

$[(1,0,0,4/3),(0,1,0,-4/3),(0,0,1,10/3)]$

So we get : $(x,y,z)=(4/3,-4/3,10/3)$

(Just check my aritmetic, I tend to make mistakes)

How do you solve using gaussian elimination or gauss-jordan elimination, 2x + y - z = -22x+y−z=−22x + y - z = -2, x + 3y + 2z = 4x+3y+2z=4x + 3y + 2z = 4, 3x + 3y - 3z = -103x+3y−3z=−103x + 3y - 3z = -10?