How do you solve using gaussian elimination or gauss-jordan elimination, $2 x - y + 5 z - t = 7$ $2x - y + 5z - t = 7$ , $x + 2 y - 3 t = 6$ $x + 2y - 3t = 6$ , $3 x - 4 y + 10 z + t = 8$ $3x - 4y + 10z + t = 8$ ?

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Answer 1 · 2017-08-12T09:57:55

The solutions are $((x),(y),(z),(t))=((4-2z+t),(1+z+t),(z),(t))$

We perform the Gauss Jordan elimination with the augmented matrix

$((1,2,0,-3,:,6),(2,-1,5,-1,:,7),(3,-4,10,1,:,8))$

$R3larrR3-3R1$ , $=>$ , $((1,2,0,-3,:,6),(2,-1,5,-1,:,7),(0,-10,10,10,:,-10))$

$R2larrR2-2R1$ , $=>$ , $((1,2,0,-3,:,6),(0,-5,5,5,:,-5),(0,-10,10,10,:,-10))$

$R3larr(R3)/(-10)$ , $=>$ , $((1,2,0,-3,:,6),(0,-5,5,5,:,-5),(0,1,-1,-1,:,1))$

$R2larr(R2)/(-5)$ , $=>$ , $((1,2,0,-3,:,6),(0,1,-1,-1,:,1),(0,1,-1,-1,:,1))$

$R3larrR3-R2$ , $=>$ , $((1,2,0,-3,:,6),(0,1,-1,-1,:,1),(0,0,0,0,:,0))$

$R1larrR1-2R2$ , $=>$ , $((1,0,2,-1,:,4),(0,1,-1,-1,:,1),(0,0,0,0,:,0))$

The solutions are

$((x),(y),(z),(t))=((4-2z+t),(1+z+t),(z),(t))$

$z$ and $t$ are free

How do you solve using gaussian elimination or gauss-jordan elimination, 2x - y + 5z - t = 72x−y+5z−t=72x - y + 5z - t = 7, x + 2y - 3t = 6x+2y−3t=6x + 2y - 3t = 6, 3x - 4y + 10z + t = 83x−4y+10z+t=83x - 4y + 10z + t = 8?