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

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Answer 1 · 2016-11-15T17:14:43

Please see the explanation for steps leading to the solution:

$x = - 3, y = \frac{1}{2}, and z = 1$ $x = -3, y = 1/2, and z = 1$

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

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