How do you solve the system #x-2y+2z=-2, -x+4y-2z=3, 2x-4y+z=-7#?

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Answer 1 · 2016-11-15T17:14:43

Please see the explanation for steps leading to the solution:

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

Write the 3 equations as an augmented matrix:

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

Add row 1 to row 2:

#[ (1,-2,2,|,-2), (0,2,0,|,1), (2,-4,1,|,-7) ]#

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

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

Divide row 3 by -3:

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

Add row 2 to row 1:

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

Divide row 2 by 2:

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

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

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

The means that #x = -3, y = 1/2, and z = 1#

check:

#(-3) - 2(1/2) + 2(1) = -2#
#-(-3) + 4(1/2) -2(1) = 3#
#2(-3) - 4(1/2) + 1 = -7#

#-2 = -2#
#3 = 3#
#-7 = -7#

This checks