How do you solve the system #5x-4y+2z=21#, #-x-5y+6z=-24#, and #-x-4y+5z=-21#?

1 Answer
Oct 11, 2016

Answer:

Please see the explanation for the matrix row operations.

#x = 5, y = -1# and #z = -4#

Explanation:

Write the row for −x − 5y + 6z = −24 into an augmented matrix:

#[ (-1, -5, 6, |, -24) ]#

Add the row for −x − 4y + 5z = −21:

#[ (-1, -5, 6, |, -24), (-1, -4, 5, |, -21) ]#

Add the row for 5x − 4y + 2z = 21:

#[ (-1, -5, 6, |, -24), (-1, -4, 5, |, -21), (5, -4, 2, |, 21) ]#

Multiply row 1 by -1 (and leave it that way) then add to row 2:

#[ (1, 5, -6, |, 24), (0, 1, -1, |, 3), (5, -4, 2, |, 21) ]#

Multiply row 1 by -5 and add to row 3:

#[ (1, 5, -6, |, 24), (0, 1, -1, |, 3), (0, -29, 32, |, -99) ]#

Multiply row 2 by 29 and add to row 3:

#[ (1, 5, -6, |, 24), (0, 1, -1, |, 3), (0, 0, 3, |, -12) ]#

Divide row 3 by 3:

#[ (1, 5, -6, |, 24), (0, 1, -1, |, 3), (0, 0, 1, |, -4) ]#

Add row 3 to row 2:

#[ (1, 5, -6, |, 24), (0, 1, 0, |, -1), (0, 0, 1, |, -4) ]#

Multiply row 3 by 6 and add to row 1:

#[ (1, 5, 0, |, 0), (0, 1, 0, |, -1), (0, 0, 1, |, -4) ]#

Multiply row 2 by -5 and add to row 1:

#[ (1, 0, 0, |, 5), (0, 1, 0, |, -1), (0, 0, 1, |, -4) ]#

This means that #x = 5, y = -1# and #z = -4#