How do you solve the below equation using reduced row-echelon form?

#2x+3y-4z=9#
#x+y+z=3#
#2x+y-3z=2#

1 Answer
Feb 25, 2018

Answer:

#x = -7/11,y = 39/11, z = 1/11#

Explanation:

Start with the augmented matrix #(A|b)#

#((2,3,-4,|,9),(1,1,1,|,3),(2,1,-3,|,2))#

We carry out row reductions to transform this to the reduced row-echelon form. We first swap rows 1 and 2 (this is not necessary - but will cut down some of the computation)

#((1,1,1,|,3),(2,3,-4,|,9),(2,1,-3,|,2))#

#R_2 -> R_2-2R_1, R_3->R_3-2R_1#
#((1,1,1,|,3),(0,1,-6,|,3),(0,-1,-5,|,-4))#

#R_1->R_1-R_2, R_3->R_3+R_2#
#((1,0,7,|,0),(0,1,-6,|,3),(0,0,-11,|,-1))#

#R_3 -> R_3/{-11}#
#((1,0,7,|,0),(0,1,-6,|,3),(0,0,1,|,1/11))#

#R_1 -> R_1-7R_3, R_2 -> R_2+6R_3#
#((1,0,0,|,-7/11),(0,1,0,|,39/11),(0,0,1,|,1/11))#

This gives
#x = -7/11,y = 39/11, z = 1/11#