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

Answer:

#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)#