How do you solve using gaussian elimination or gauss-jordan elimination, $2x + 2y - 3z = -2$ , $3x - 1 - 2z = 1$ , $2x + 3y - 5z = -3$ ?

Question

2134 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-06T19:35:01

$P=[(-2, 1, 0)]$

I'm assuming that in the second row is missing y. Solving this matrix:

$([2,2,-3,|,-2],[3,-1,-2,|,1],[2,3,-5,|,-3])$
$R_1=R_1xx3$
$R_2=R_2xx-2$
$R_3=R_3xx-3$

$([6,6,-9,|,-6],[-6,2,4,|,-2],[-6,-9,15,|,9])$
$R_2=R_2+R_1$
$R_3=R_3+R_1$

$([6,6,-9,|,-6],[0,8,-5,|,-8],[0,-3,6,|,3])$
$R_2=R_2xx3$
$R_3=R_3xx8$

$([6,6,-9,|,-6],[0,24,-15,|,-24],[0,-24,48,|,24])$
$R_3=R_3+R_2$
$R_1=R_1xx1/3$

$([2,2,-3,|,-2],[0,24,-15,|,-24],[0,0,33,|,0])$
$R_2=R_2xx1/3$
$R_3=R_3xx5/33$

$([2,2,-3,|,-2],[0,8,-5,|,8],[0,0,5,|,0])$
$R_2=R_2+R_3$
$R_3=R_3xx1/5$

$([2,2,-3,|,-2],[0,8,0,|,8],[0,0,1,|,0])$
$R_2=R_2xx1/8$
$R_1=R_1-2xxR_2$
$R_1=R_1+3xxR_3$

$([2,0,0,|,-4],[0,1,0,|,1],[0,0,1,|,0])$
$R_1=R_1xx1/2$

$([1,0,0,|,-2],[0,1,0,|,1],[0,0,1,|,0])$

$P=[(-2, 1, 0)]$

(too long)

How do you solve using gaussian elimination or gauss-jordan elimination, 2x + 2y - 3z = -22x + 2y - 3z = -2, 3x - 1 - 2z = 13x - 1 - 2z = 1, 2x + 3y - 5z = -32x + 3y - 5z = -3?