How do you solve the system of equations #x+ 4y - 2z = 11, 2x + 9y - 7z = - 3, - 3x + y - z = - 6#?

1 Answer
Douglas K.
Dec 16, 2016

Write the coefficients of all 3 equations into an augmented matrix. Perform row operations until you obtain an identity matrix. Please see the explanation.

Explanation:

Write the coefficients of all 3 equations into an augmented matrix:

#x + 4y - 2z = 11 to [(1, 4,-2,|,11)]#

#2x + 9y - 7z = -3 to [(1, 4,-2,|,11),(2,9,-7,|,-3)]#

#-3x + y - z = -6 to [(1, 4,-2,|,11),(2,9,-7,|,-3),(-3,1,-1,|,-6)]#

Perform row operations.

#[ (1, 4,-2,|,11), (2,9,-7,|,-3), (-3,1,-1,|,-6) ]#

#-2R_1 + R_2 to R_2#

#[ (1, 4,-2,|,11), (0,1,-3,|,-25), (-3,1,-1,|,-6) ]#

#3R_1 + R_3 to R_3#

#[ (1, 4,-2,|,11), (0,1,-3,|,-25), (0,13,-7,|,27) ]#

#-13R_2 + R_3 to R_3#

#[ (1, 4,-2,|,11), (0,1,-3,|,-25), (0,0,32,|,352) ]#

#R_3/32 to R_3#

#[ (1, 4,-2,|,11), (0,1,-3,|,-25), (0,0,1,|,11) ]#

#3R_3 + R_2 to R_2#

#[ (1, 4,-2,|,11), (0,1,0,|,8), (0,0,1,|,11) ]#

#2R_3 + R_1 to R_1#

#[ (1, 4,0,|,33), (0,1,0,|,8), (0,0,1,|,11) ]#

#-4R_2 + R_1 to R_1#

#[ (1, 0,0,|,1), (0,1,0,|,8), (0,0,1,|,11) ]#

#x = 1, y = 8 and z = 11#

Check:

#1 + 4(8) - 2(11) = 11#

#2(1) + 9(8) - 7(11) = -3#

#-3(1) + (8) - 11 = -6#

#11 = 11#

#-3 = -3#

#-6 = -6#

This checks.

Related questions
Impact of this question
520 views around the world
You can reuse this answer
Creative Commons License