How do you solve using gaussian elimination or gauss-jordan elimination, $x_{1} + 3 x_{2} + x_{3} + x_{4} = 3$ $x_1 + 3x_2 +x_3 + x_4= 3$ , $2 x_{1} - 2 x_{2} + x_{3} + 2 x_{4} = 8$ $2x_1- 2x_2 + x_3 + 2x_4 =8$ and $3 x_{1} + x_{2} + 2 x_{3} - x_{4} = - 1$ $3x_1 + x_2 + 2x_3 - x_4 =-1$ ?

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Answer 1 · 2018-01-07T09:37:09

$P={((-6-5p)/8, (2-p)/8, p, 3), p in RR}$

$([1,3,1,1,|,3],[2,-2,1,2,|,8],[3,1,2,-1,|,-1])~~([1,3,1,1,|,3],[0,-8,-1,0,|,2],[0,-8,-1,-4,|,-10])$
$R_2=R_2-2xxR_1$
$R_3=R_3-3xxR_1$
$R_3=R_3-R_2$

$([1,3,1,1,|,3],[0,-8,-1,0,|,2],[0,0,0,-4,|,-12])~~([1,3,1,1,|,3],[0,-8,-1,0,|,2],[0,0,0,1,|,3])$
$R_3=R_3xx-1/4$
$R_1=R_1-R_3$

$([1,3,1,0,|,0],[0,-8,-1,0,|,2],[0,0,0,1,|,3])~~([1,3,1,0,|,0],[0,1,1/8,0,|,1/4],[0,0,0,1,|,3])$
$R_2=R_2xx-1/8$
$R_1=R_1-3xxR_2$

$([1,0,5/8,0,|,-3/4],[0,1,1/8,0,|,1/4],[0,0,0,1,|,3])$

$x_3=pcolor(white)(QQQQQQQ)p in RR$ (as a parameter)

$=>x_1=-3/4-5/8p=(-6-5p)/8$

$=>x_2=1/4-1/8p=(2-p)/8$

$=>x_4=3$

$P={((-6-5p)/8, (2-p)/8, p, 3), p in RR}$

How do you solve using gaussian elimination or gauss-jordan elimination, x_1 + 3x_2 +x_3 + x_4= 3x1+3x2+x3+x4=3x_1 + 3x_2 +x_3 + x_4= 3, 2x_1- 2x_2 + x_3 + 2x_4 =82x1−2x2+x3+2x4=82x_1- 2x_2 + x_3 + 2x_4 =8 and 3x_1 + x_2 + 2x_3 - x_4 =-13x1+x2+2x3−x4=−13x_1 + x_2 + 2x_3 - x_4 =-1?