How do you solve using gaussian elimination or gauss-jordan elimination, $2x + 4y−6z = 42$ , $x + 2y+ 3z = 3$ , $3x−4y+ 4z = −16$ ?

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Answer 1 · 2017-11-20T12:50:06

The solution is $((x),(y),(z))=((4),(4),(-3))$

Perform the Gauss Jordan elimination on the augmented matrix

$A=((1,2,-3,|,21),(1,2,3,|,3),(3,-4,4,|,-16))$

Perform the row operations

$R2larrR2-R1$ and $R3larrR3-3R1$

$A=((1,2,-3,|,21),(0,0,6,|,-18),(0,-10,13,|,-79))$

$R2harrR3$ and $R3larr(R3)/6$

$A=((1,2,-3,|,21),(0,-10,13,|,-79),(0,0,1,|,-3))$

$R2larr(R2-13R3)$ and $R2larr(R2)/(-10)$

$A=((1,2,-3,|,21),(0,1,0,|,4),(0,0,1,|,-3))$

$R1larr(R1-2R2)$ and $R1larrR1+3R3$

$A=((1,0,0,|,4),(0,1,0,|,4),(0,0,1,|,-3))$

How do you solve using gaussian elimination or gauss-jordan elimination, 2x + 4y−6z = 422x + 4y−6z = 42, x + 2y+ 3z = 3x + 2y+ 3z = 3, 3x−4y+ 4z = −163x−4y+ 4z = −16?