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

1 Answer
Jan 7, 2018

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

Explanation:

([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}