Answer:
The solutions are
.
Explanation:
To find the solution of the the system of linear equations represented by this augmented matrix
![\left[ \begin{array}{cccc} 2 & 1 & -1 & 3 \\\\ 1 & -1 & 1 & 0 \\\\ 0 & 1 & 2 & 1 \end{array} \right]](https://img.qammunity.org/2021/formulas/mathematics/college/x02pu2x9dshg7mouhpovpo49b222i2qe5w.png)
First, transform the augmented matrix to the reduced row echelon form. Any matrix can be transformed into its echelon forms, using a series of elementary row operations.
There are three kinds of elementary matrix operations.
- Interchange two rows (or columns).
- Multiply each element in a row (or column) by a non-zero number.
- Multiply a row (or column) by a non-zero number and add the result to another row (or column).
Applying the following elementary matrix operations:
Row Operation 1: Divide row 1 by 2

Row Operation 2: Subtract row 1 from row 2

Row Operation 3: Multiply row 2 by −2/3

Row Operation 4: Subtract row 2 multiplied by 1/2 from row 1

Row Operation 5: Subtract row 2 from row 3

Row Operation 6: Divide row 3 by 3

Row Operation 7: Add row 3 to row 2

We get that the reduced row echelon form of the augmented matrix is:
![\left[ \begin{array}{cccc} 1 & 0 & 0 & 1 \\\\ 0 & 1 & 0 & 1 \\\\ 0 & 0 & 1 & 0 \end{array} \right]](https://img.qammunity.org/2021/formulas/mathematics/college/e4a6gqr9aro4agnvt9o8835z0jfmf3fmx8.png)
which corresponds to the system

Since every column in the coefficient part of the matrix has a leading entry that means our system has a unique solution:
