A system of linear equations can be solved in many ways, elimination, substitution and a method using matrices known as Gauss–Jordan elimination. Which of these is the easiest method is dependent on the system of equations that is being solved. For a system consisting of a large number of equations, the Gauss–Jordan elimination method is the easiest to use especially with the help of a computer algorithm.

For the given system of equations, the initial matrix that can be created is:

`[[1,3,2|18],[1,-1,1|4],[1,1,1|10]]`

This has to be converted to the reduced echelon form. The operations required to accomplish that are:

`[[1,3,2|18],[1,-1,1|4],[1,1,1|10]]`

Subtact row 3 from row 2

=> `[[1,3,2|18],[0,-2,0|-6],[1,1,1|10]]`

Divide row 2 by -2

=>  `[[1,3,2|18],[0,1,0|3],[1,1,1|10]]`

Subtract 3 times row 2 from row 1

=> `[[1,0,2|9],[0,1,0|3],[1,1,1|10]]`

Subtract row 1 and row 2 from row 3

=> `[[1,0,2|9],[0,1,0|3],[0,0,-1|-2]]`

Divide row 3 by -1

=>  `[[1,0,2|9],[0,1,0|3],[0,0,1|2]]`

Subtract 2 times row 3 from row 1

=> `[[1,0,0|5],[0,1,0|3],[0,0,1|2]]`

The matrix is now in reduced row echelon form.

The solution of the given set of linear equations is x = 5, y = 3 and z = 2