To use the matrix reduced row echelon form (RREF) for the system of equations:

1. Write the augmented matrix for the system of equations. For your equations $2x - y = 2$ and $-3x + 2y = -4$, the augmented matrix looks like this:

   2 & -1 & | & 2 \\ -3 & 2 & | & -4 \end{bmatrix}$$

2. Perform row operations to get the matrix into RREF.

   • First, we can multiply the first row by $\frac{1}{2}$:
   1 & -\frac{1}{2} & | & 1 \\ -3 & 2 & | & -4 \end{bmatrix}$$ - Next, eliminate the first element in the second row by adding $$3 \times R_1$$ to $$R_2$$: $$R_2 \rightarrow R_2 + 3R_1 \Rightarrow \begin{bmatrix} 1 & -\frac{1}{2} & | & 1 \\ 0 & \frac{1}{2} & | & -1 \end{bmatrix}$$ - Now, multiply the second row by $$2$$: $$R_2 \rightarrow 2R_2 \Rightarrow \begin{bmatrix} 1 & -\frac{1}{2} & | & 1 \\ 0 & 1 & | & -2 \end{bmatrix}$$ - Finally, eliminate the second element in the first row by adding $$\frac{1}{2}R_2$$ to $$R_1$$: $$R_1 \rightarrow R_1 + \frac{1}{2} R_2 \Rightarrow \begin{bmatrix} 1 & 0 & | & 0 \\ 0 & 1 & | & -2 \end{bmatrix}$$

3. Interpret the RREF: The final matrix represents the equations:

   y = -2$$

Thus, the solution to the system is $x = 0$ and $y = -2$.

Would you like more details on any step? Here are some questions to consider:

1. What are row operations in detail?
2. How do you interpret the solution from RREF?
3. Can RREF be used for larger systems of equations?
4. What if the system has no solutions or infinitely many solutions?
5. How can you verify the solution using substitution?

Tip: Always double-check your row operations to ensure accuracy!