How do you solve using gaussian elimination or gauss-jordan elimination, x + y + z - 3t = 1, 2x + y + z - 5t = 0, y + z - t = 2, 3x - 2z + 2t = -7#?

1 Answer
Jan 4, 2018

P=({-1+2t,quad-3t,quad2+4t}, t in RR)

Explanation:

( [1,1,1,-3,|,1],[2,1,1,-5,|,0],[0,1,1,-1,|,2],[3,0,-2,2,|,-7] )
R_2=R_2-2xxR_1
R_4=R_4-3xxR_1

( [1,1,1,-3,|,1],[2-2,1-2,1-2,-5+6,|,0-2],[0,1,1,-1,|,2],[3-3,0-3,-2-3,2+9,|,-7-3] )~~( [1,1,1,-3,|,1],[0,-1,-1,1,|,-2],[0,1,1,-1,|,2],[0,-3,-5,11,|,-10] )
R_3=R_3+R_2
R_4=R_4-3xxR_2

( [1,1,1,-3,|,1],[0,-1,-1,1,|,-2],[0,1-1,1-1,-1+1,|,2-2],[0,-3+3,-5+3,11-3,|,-10+6] )~~( [1,1,1,-3,|,1],[0,-1,-1,1,|,-2],[0,0,0,0,|,0],[0,0,-2,8,|,-4] )~
R_4=-1/2xxR_4
R_4hArrR_3
R_2=-1xxR_2

( [1,1,1,-3,|,1],[0,1,1,-1,|,2],[0,0,1,-4,|,2],[0,0,0,0,|,0] )
R_2=R_2-R_3

( [1,1,1,-3,|,1],[0,1,0,3,|,0],[0,0,1,-4,|,2],[0,0,0,0,|,0] )
R_1=R_1-R_2
R_1=R_1-R_3

( [1,1-1,1-1,-3-3+4,|,1-2],[0,1,0,3,|,0],[0,0,1,-4,|,2],[0,0,0,0,|,0] )~~( [1,0,0,-2,|,-1],[0,1,0,3,|,0],[0,0,1,-4,|,2],[0,0,0,0,|,0] )


x-2t=-1
y+3t=0
z-4t=2

Let t be a parameter, t in RR

x=-1+2t
y=-3t
z=2+4t

P=({-1+2t,quad-3t,quad2+4t}, t in RR)