1. **Problem:** Determine the solution type for the system with reduced row echelon form (RREF): $$\begin{bmatrix}1 & 0 & 0 & | & -2 \\ 0 & 1 & 0 & | & 3 \\ 0 & 0 & 0 & | & 0\end{bmatrix}$$ 2. **Formula and rules:** In RREF, each leading 1 corresponds to a pivot variable. If a row has all zeros except the augmented part is nonzero, the system has no solution. If there are free variables (columns without pivots), the system has infinitely many solutions. Otherwise, the system has a unique solution. 3. **Analysis:** Here, pivots are in columns 1 and 2 (variables $x$ and $y$). The third variable $z$ is free since column 3 has no pivot. 4. **Interpretation:** The last row is all zeros, so no contradiction. Since $z$ is free, infinitely many solutions exist. 5. **Solution form:** $$x = -2, \quad y = 3, \quad z = t \quad \text{where } t \in \mathbb{R}$$ --- 1. **Problem:** Determine the solution type for the system with RREF: $$\begin{bmatrix}1 & 0 & 0 & | & 0 \\ 0 & 1 & 0 & | & 1 \\ 0 & 0 & 1 & | & -2\end{bmatrix}$$ 2. **Formula and rules:** Each pivot corresponds to a unique variable solution. No free variables means unique solution. 3. **Analysis:** Pivots in columns 1, 2, and 3 correspond to $x$, $y$, and $z$ respectively. 4. **Solution:** $$x = 0, \quad y = 1, \quad z = -2$$ --- 1. **Problem:** Determine the solution type for the system with RREF: $$\begin{bmatrix}1 & 0 & | & 0 \\ 0 & 1 & | & 0 \\ 0 & 0 & | & 0\end{bmatrix}$$ 2. **Formula and rules:** The system has two variables $x$ and $y$ with pivots in both columns, and the last row is zero. 3. **Analysis:** No contradictions, no free variables. 4. **Solution:** Unique solution $$x = 0, \quad y = 0$$ --- 1. **Problem:** Determine the solution type for the system with RREF: $$\begin{bmatrix}1 & 0 & 0 & | & -2 \\ 0 & 1 & 0 & | & 3 \\ 0 & 0 & 0 & | & 3\end{bmatrix}$$ 2. **Formula and rules:** The last row corresponds to $0x + 0y + 0z = 3$, which is impossible. 3. **Conclusion:** The system is inconsistent and has no solutions. --- **Final answers:** 1. Infinitely many solutions with $x = -2$, $y = 3$, $z$ free. 2. Unique solution: $x=0$, $y=1$, $z=-2$. 3. Unique solution: $x=0$, $y=0$. 4. No solutions.