How do you solve these set of linear equations: #-2x + y - z = 2; - x - 3y + z = - 10; 3x + 6z = - 24#?

I use an augmented matrix of coefficients and perform row operations on the matrix:

For the first row, I will write the coefficients for the equation #−x −3y +z = −10#:

|-1 -3 1|-10|

For the second row, I will write the coefficients for the equation #-2x + y - z = 2#

|-1 -3 1|-10|
|-2 1 -1|2|

For the third row, I will write the coefficients for the equation #3x + 6z = -24#

|-1 -3 1|-10|
|-2 1 -1|2|
|3 0 6|-24|

Multiply the first row by -1:

|1 3 -1|10|
|-2 1 -1|2|
|3 0 6|-24|

Multiply the first row by 2 and add to the second row::

|1 3 -1|10|
|0 7 -3|22|
|3 0 6|-24|

Multiply the first row by -3 and add to the third row::

|1 3 -1|10|
|0 7 -3|22|
|0 -9 9|-54|

Divide the third row by -9:

|1 3 -1|10|
|0 7 -3|22|
|0 1 -1|6| (Edit: correct the third column from 1 to -1

Interchange rows 2 and 3:

|1 3 -1|10|
|0 1 -1|6|
|0 7 -3|22|

Multiply the second row by -7 and add to the third row:

|1 3 -1|10|
|0 1 1|6|
|0 0 4|-20|

Divide the third row by 4:

|1 3 -1|10|
|0 1 1|6|
|0 0 1|-5|

Subtract the third two from the second row:

|1 3 -1|10|
|0 1 0|1|
|0 0 1|-5|

Add the third two to the first row:

|1 3 0|5|
|0 1 0|1|
|0 0 1|-5|

Multiply the second row by - 3 and add to the first row:

|1 0 0|2|
|0 1 0|1|
|0 0 1|-5|

We know that we are done, because the main diagonal of the left side all 1s and there are all 0s, elsewhere.

This means x = 2, y = 1 and z = -5.