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

##### 1 Answer

Please see the explantion

#### Explanation:

Write

Add a row for the equation

Add a row for the equation

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

We want the coefficient in position

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

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:

#2y+z=3

#y=3/2-z/2

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