Question

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

Answer 1

Please see the explanation for steps leading to the solution:

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

Explanation:

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