Assume that the augmented matrix of the system of linear equations is given as follows:
$$
\begin{bmatrix}
2 & 1 & -3 & 2 \\
2 & -3 & 1 & 10 \\
-2 & 1 & 1 & -6
\end{bmatrix}
$$
Use the following steps to reduce the augmented matrix to generalized row echelon form:
Step 1:
$$R_3 + R_1 \rightarrow R_3$$
$$R_2 - R_1 \rightarrow R_2$$
Step 2:
$$(-\frac{1}{4})R_2 \rightarrow R_2$$
$$(-\frac{1}{2})R_3 \rightarrow R_3$$
Step 3:
$$R_3 + R_2 \rightarrow R_3$$
Deduce that the solutions are:
$$x = t+1, y = t-2, z = t \text{ with } t \in \mathbb{R}$$
$$x = t, y = t+2, z = t \text{ with } t \in \mathbb{R}$$
$$x = t+2, y = t-2, z = t \text{ with } t \in \mathbb{R}$$
$$x = -2, y = t-2, z = t \text{ with } t \in \mathbb{R}$$
None of the given options
