Question

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

Answer 1

Please see the explantion

Explanation:

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

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

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

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

Add a row for the equation `-3x-2y+5z=-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 `(1,1)` 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_1toR_2`

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

`R_3+3R_1toR_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-2z=3`
#2y+z=3

`z= x/2-3/2`
#y=3/2-z/2

`z= x/2-3/2" [1]"`
`y=9/4-x/4" [2]"`

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