How do you solve using gaussian elimination or gauss-jordan elimination, $3 y + 2 z = 4$ $3y + 2z = 4$ , $2 x - y - 3 z = 3$ $2x − y − 3z = 3$ , $2 x + 2 y - z = 7$ $2x+ 2y − z = 7$ ?

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Answer 1 · 2017-12-23T20:04:59

${((13+7p)/6, (4-2p)/3, p), p in RR}$

$0x+3y+2z=4$
$2x−y−3z=3$
$2x+2y−z=7$

$((0,3,2quadquad|4),(2,-1,-3|3),(2,2,-1|7))~~((2,-1,-3|3),(2,2,-1|7),(0,3,2quadquad|4))~~$
$R_3hArrR_1$
$R_2=R_2-R_1$

$x-7/6z=13/6$

$y+2/3z=4/3$

Last row has all zeros which means we have to substitute for parameter. z seems to be the best way to go.

$z=p$ (p as parameter)

$x-7/6p=13/6quad=>quadx=13/6+7/6p=(13+7p)/6$

$y+2/3p=4/3quad=>quady=4/3-2/3p=(4-2p)/3$

${((13+7p)/6, (4-2p)/3, p), p in RR}$

How do you solve using gaussian elimination or gauss-jordan elimination, 3y + 2z = 43y+2z=43y + 2z = 4, 2x − y − 3z = 32x−y−3z=32x − y − 3z = 3, 2x+ 2y − z = 72x+2y−z=72x+ 2y − z = 7?