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

1 Answer
Jan 6, 2018

Answer:

#P=[(-2, 1, 0)]#

Explanation:

I'm assuming that in the second row is missing y. Solving this matrix:

#([2,2,-3,|,-2],[3,-1,-2,|,1],[2,3,-5,|,-3])#
#R_1=R_1xx3#
#R_2=R_2xx-2#
#R_3=R_3xx-3#

#([6,6,-9,|,-6],[-6,2,4,|,-2],[-6,-9,15,|,9])#
#R_2=R_2+R_1#
#R_3=R_3+R_1#

#([6,6,-9,|,-6],[0,8,-5,|,-8],[0,-3,6,|,3])#
#R_2=R_2xx3#
#R_3=R_3xx8#

#([6,6,-9,|,-6],[0,24,-15,|,-24],[0,-24,48,|,24])#
#R_3=R_3+R_2#
#R_1=R_1xx1/3#

#([2,2,-3,|,-2],[0,24,-15,|,-24],[0,0,33,|,0])#
#R_2=R_2xx1/3#
#R_3=R_3xx5/33#

#([2,2,-3,|,-2],[0,8,-5,|,8],[0,0,5,|,0])#
#R_2=R_2+R_3#
#R_3=R_3xx1/5#

#([2,2,-3,|,-2],[0,8,0,|,8],[0,0,1,|,0])#
#R_2=R_2xx1/8#
#R_1=R_1-2xxR_2#
#R_1=R_1+3xxR_3#

#([2,0,0,|,-4],[0,1,0,|,1],[0,0,1,|,0])#
#R_1=R_1xx1/2#

#([1,0,0,|,-2],[0,1,0,|,1],[0,0,1,|,0])#

#P=[(-2, 1, 0)]#

(too long)