How do you solve the system #x+y-z=2#, #x-2z=1#, and #2x-3y-z=8#?

1 Answer
Dec 20, 2016

Answer:

Write the coefficients of the equations into an augmented matrix and then perform Gauss-Jordan Elimination, using row operations.

Explanation:

Write the coefficients of the equations into an augmented matrix:

#x - 2z = 1 to [(1, 0, -2,|,1)]#

#x + y - z = 2 to [(1, 0, -2,|,1),(1, 1, -1,|,2)]#

#2x - 3y - z = 8 to [(1, 0, -2,|,1),(1, 1, -1,|,2),(2, -3, -1,|,8)]#

Perform Guass-Jordan Elimination using row operations on the augmented matrix:

#[ (1, 0, -2,|,1), (1, 1, -1,|,2), (2, -3, -1,|,8) ]#

Elementary Row Operations Notation

#R_2 - R_1 to R_2#

#[ (1, 0, -2,|,1), (0, 1, 1,|,1), (2, -3, -1,|,8) ]#

#R_3 - 2R_1 to R_3#

#[ (1, 0, -2,|,1), (0, 1, 1,|,1), (0, -3, 3,|,6) ]#

#3R_2 + R_3 to R_3#

#[ (1, 0, -2,|,1), (0, 1, 1,|,1), (0, 0, 6,|,9) ]#

#R_3/6#

#[ (1, 0, -2,|,1), (0, 1, 1,|,1), (0, 0, 1,|,3/2) ]#

#R_2 - R_3 to R_2#

#[ (1, 0, -2,|,1), (0, 1, 0,|,-1/2), (0, 0, 1,|,3/2) ]#

#2R_3 + R_1 to R_1#

#[ (1, 0, 0,|,4), (0, 1, 0,|,-1/2), (0, 0, 1,|,3/2) ]#

#x = 4, y = -1/2, and z = 3/2#

check:

#4 - 2(3/2) = 1#
#4 + (-1/2) - 3/2 = 2#
#2(4) - 3(-1/2) - 3/2 = 8#

#1 = 1#
#2 = 2#
#8 = 8#

This checks.