How do you solve the system #x-6y+4z=-12#, #x+y-4z=12#, and #2x+2y+5z=-15#?

1 Answer
Nov 10, 2016

Answer:

Please see the explanation for steps leading to the answer:

#x = 0, y = 0, and z = -3#

Explanation:

Write the 3 equations as an augmented matrix:

#[ (1,-6,4,|,-12), (1,1,-4,|,12), (2,2,5,|,-15) ]#

Subtract row 1 from row 2:

#[ (1,-6,4,|,-12), (0,7,-8,|,24), (2,2,5,|,-15) ]#

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

#[ (1,-6,4,|,-12), (0,7,-8,|,24), (0,14,-3,|,9) ]#

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

#[ (1,-6,4,|,-12), (0,7,-8,|,24), (0,0,13,|,-39) ]#

Divide row 3 by 13:

#[ (1,-6,4,|,-12), (0,7,-8,|,24), (0,0,1,|,-3) ]#

Multiply row 3 by 8 and add to row 2:

#[ (1,-6,4,|,-12), (0,7,0,|,0), (0,0,1,|,-3) ]#

Multiply row 3 by -4 and add to row 1:

#[ (1,-6,0,|,0), (0,7,0,|,0), (0,0,1,|,-3) ]#

Divide row 2 by 7:

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

Multiply row 2 by 6 and add to row 1:

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

#x = 0, y = 0, and z = -3#