How do you solve the system #2x+y+z=0, 3x-2y-3z=-21, 4x+5y+3z=-2#?

1 Answer
Nov 18, 2016

Answer:

Please see the explanation.

Explanation:

Write the equations as an augmented matrix:

#| (2,1,1,|,0), (3, -2,-3,|,-21), (4,5,3,|,-2) |#

Multiply row 2 by -2:

#| (2,1,1,|,0), (-6, 4,6,|,42), (4,5,3,|,-2) |#

Multiply row 1 by 3 and add to row 2:

#| (2,1,1,|,0), (0, 7,9,|,42), (4,5,3,|,-2) |#

Multiply row 1 by -2 and add to row 3:

#| (2,1,1,|,0), (0, 7,9,|,42), (0,3,1,|,-2) |#

Multiply row 3 by -2 and add to row 2:

#| (2,1,1,|,0), (0, 1,7,|,46), (0,3,1,|,-2) |#

Multiply row 2 by -3 and add to row 3:

#| (2,1,1,|,0), (0, 1,7,|,46), (0,0,-20,|,-140) |#

Multiply row 2 by 7/20

#| (2,1,1,|,0), (0, 1,7,|,46), (0,0,-7,|,-49) |#

Add row 3 to row 2:

#| (2,1,1,|,0), (0, 1,0,|,-3), (0,0,-7,|,-49) |#

Divide row 3 by -7

#| (2,1,1,|,0), (0, 1,0,|,-3), (0,0,1,|,7) |#

Subtract row 3 from row 1:

#| (2,1,0,|,-7), (0, 1,0,|,-3), (0,0,1,|,7) |#

Subtract row 2 from row 1:

#| (2,0,0,|,-4), (0, 1,0,|,-3), (0,0,1,|,7) |#

Divide row 1 by 2:

#| (1,0,0,|,-2), (0, 1,0,|,-3), (0,0,1,|,7) |#

#x = -2, y = -3, and z = 7#

check:

#2(-2) + (-3) + 7 = 0#
#3(-2) - 2(-3) -3(7) = -21#
#4(-2) + 5(-3) + 3(7) = -2#

#0 = 0#
#-21 = -21#
#-2 = -2#

This checks.