How do you solve using gaussian elimination or gauss-jordan elimination, $2x - y + 3z = 24$ , $2y - z = 14$ , $7x - 5y = 6$ ?

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Answer 1 · 2017-09-01T15:41:25

The solution is $((x),(y),(z))=((8),(10),(6))$

The augmented matrix is

$((2,-1,3,:,24),(0,2,-1,:,14),(7,-5,0,:,6))$

We can perform the Gauss-Jordan elimination

$R3 harr R2$ , $=>$ , $((2,-1,3,:,24),(7,-5,0,:,6),(0,2,-1,:,14))$

$R2larrR2-3R1$ , $=>$ , $((2,-1,3,:,24),(1, -2,-9,:,-66),(0,2,-1,:,14))$

$R2larr2R2-R1$ , $=>$ , $((2,-1,3,:,24),(0, -3,-21,:,-156),(0,2,-1,:,14))$

$R3larr3R3+2R2$ , $=>$ , $((2,-1,3,:,24),(0, -3,-21,:,-156),(0,0,-45,:,-270))$

$R3larr(R3)/(-45)$ , $=>$ , $((2,-1,3,:,24),(0, -3,-21,:,-156),(0,0,1,:,6))$

$R2larrR2+21R3$ , $=>$ , $((2,-1,3,:,24),(0, -3,0,:,-30),(0,0,1,:,6))$

$R2larr(R2)/(-3)$ , $=>$ , $((2,-1,3,:,24),(0, 1,0,:,10),(0,0,1,:,6))$

$R1larrR1+R2$ , $=>$ , $((2,0,3,:,34),(0, 1,0,:,10),(0,0,1,:,6))$

$R1larrR1-3R3$ , $=>$ , $((2,0,0,:,16),(0, 1,0,:,10),(0,0,1,:,6))$

$R1larr(R1)/2$ , $=>$ , $((1,0,0,:,8),(0, 1,0,:,10),(0,0,1,:,6))$

How do you solve using gaussian elimination or gauss-jordan elimination, 2x - y + 3z = 24 2x - y + 3z = 24, 2y - z = 142y - z = 14, 7x - 5y = 67x - 5y = 6?