Question

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

Answer 1

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.