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

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Answer 1 · 2018-07-03T16:47:24

The solution is $((x),(y),(z))=((-1),(2),(1))$

Perform the Gauss-Jordan Elimination on the augmented matrix

$A=((2,1,-3,|,-3),(3,2,4,|,5),(-4,-1,2,|,4))$

Make the pivot in the first column $R1larr(R1)/2$

$((1,1/2,-3/2,|,-3/2),(3,2,4,|,5),(-4,-1,2,|,4))$

Eliminate the first column $R2larr(R2-3R1)$ and $R3larr(R3+4R1)$

$((1,1/2,-3/2,|,-3/2),(0,1/2,17/2,|,19/2),(0,1,-4,|,-2))$

Make the pivot in the $2$ nd column swap $R2harrR3$

$((1,1/2,-3/2,|,-3/2),(0,1,-4,|,-2),(0,1/2,17/2,|,19/2))$

Eliminate the second column $R3larr(R3-1/2(R2))$ , $R1larr(R1-(R2)/2)$

$((1,0,1/2,|,-1/2),(0,1,-4,|,-2),(0,0,21/2,|,21/2))$

Make the pivot in the $3rd$ column $R3larr(R3*2/21)$

$((1,0,1/2,|,-1/2),(0,1,-4,|,-2),(0,0,1,|,1))$

Eliminate the $3rd$ column $R1larr(R1-(R3)/2)$ and $R2larr(R2+4R3)$

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

The solution is

$((x),(y),(z))=((-1),(2),(1))$

How do you solve using gaussian elimination or gauss-jordan elimination, 2x + y - 3z = - 32x+y−3z=−32x + y - 3z = - 3, 3x + 2y + 4z = 53x+2y+4z=53x + 2y + 4z = 5, -4x - y + 2z = 4−4x−y+2z=4-4x - y + 2z = 4?