# How do you solve the system x+2y-z=6, -3x-2y+5z=-12, and x-2z=3?

Apr 16, 2017

#### Explanation:

Write $x - 2 z = 3$ as the first row of an Augmented Matrix :

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

Add a row for the equation $x + 2 y - z = 6$:

[ (1,0,-2,|,3), (1,2,-1,|,6) ]

Add a row for the equation $- 3 x - 2 y + 5 z = - 12$:

[ (1,0,-2,|,3), (1,2,-1,|,6), (-3,-2,5,|,-12) ]

Perform Elementary Row Operations until an identity matrix is obtained on the left.

We want the coefficient in position $\left(1 , 1\right)$ to be 1 and it is, therefore, we do not do a row operation.

We want the other two coefficients in column 1 to be 0, therefore, we perform the following 2 row operations:

${R}_{2} - {R}_{1} \to {R}_{2}$

[ (1,0,-2,|,3), (0,2,1,|,3), (-3,-2,5,|,-12) ]

${R}_{3} + 3 {R}_{1} \to {R}_{3}$

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

Because Row 3 is a scalar multiple of Row 2, one should declare that there is no unique solution.

Let's write the first two rows as linear equations:

$x - 2 z = 3$
2y+z=3

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

$z = \frac{x}{2} - \frac{3}{2} \text{ [1]}$
$y = \frac{9}{4} - \frac{x}{4} \text{ [2]}$

Equations [1] and [2] will allow you to choose a value of x and then determine the corresponding value of y and z.