Answer:
To create a matrix for this linear system, we can arrange the coefficients of the variables and the constants into a matrix as follows:
| 1 3 2 | | x | | 26 |
| 1 -3 4 | x | y | = | 2 |
| 2 1 1 | | z | | 8 |
To solve the system using row reduction, we can perform elementary row operations to transform the matrix into row echelon form or reduced row echelon form. I will use the latter approach for simplicity:
| 1 0 0 | | x | | 6 |
| 0 1 0 | x | y | = | 5 |
| 0 0 1 | | z | | -1 |
Therefore, the solution to the system is x = 6, y = 5, and z = -1.