1. **Problem statement:** Find the inverse of the 2x2 matrices $A = \begin{bmatrix} 3 & -1 \\ -1 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} -3 & 2 \\ 6 & -4 \end{bmatrix}$ using row reduction, or show that the inverse does not exist. 2. **Formula and method:** To find the inverse of a matrix $M$, we augment it with the identity matrix and perform row operations to transform $M$ into the identity matrix. The augmented part will then become $M^{-1}$. 3. **Matrix A:** Start with $\left[ \begin{array}{cc|cc} 3 & -1 & 1 & 0 \\ -1 & 4 & 0 & 1 \end{array} \right]$. Step 1: Make the leading 1 in row 1 by dividing row 1 by 3: $$\left[ \begin{array}{cc|cc} \cancel{3} \frac{1}{\cancel{3}} & -\frac{1}{3} & \frac{1}{3} & 0 \\ -1 & 4 & 0 & 1 \end{array} \right]$$ Step 2: Eliminate the $-1$ in row 2, column 1 by adding row 1 multiplied by 1 to row 2: $$\left[ \begin{array}{cc|cc} 1 & -\frac{1}{3} & \frac{1}{3} & 0 \\ 0 & 4 - \frac{1}{3} & 0 + \frac{1}{3} & 1 + 0 \end{array} \right] = \left[ \begin{array}{cc|cc} 1 & -\frac{1}{3} & \frac{1}{3} & 0 \\ 0 & \frac{12}{3} - \frac{1}{3} & \frac{1}{3} & 1 \end{array} \right] = \left[ \begin{array}{cc|cc} 1 & -\frac{1}{3} & \frac{1}{3} & 0 \\ 0 & \frac{11}{3} & \frac{1}{3} & 1 \end{array} \right]$$ Step 3: Make the leading 1 in row 2 by dividing row 2 by $\frac{11}{3}$: $$\left[ \begin{array}{cc|cc} 1 & -\frac{1}{3} & \frac{1}{3} & 0 \\ 0 & \cancel{\frac{11}{3}} \frac{3}{\cancel{11}} & \frac{1}{3} \cdot \frac{3}{11} & 1 \cdot \frac{3}{11} \end{array} \right] = \left[ \begin{array}{cc|cc} 1 & -\frac{1}{3} & \frac{1}{3} & 0 \\ 0 & 1 & \frac{1}{11} & \frac{3}{11} \end{array} \right]$$ Step 4: Eliminate the $-\frac{1}{3}$ in row 1, column 2 by adding row 2 multiplied by $\frac{1}{3}$ to row 1: $$\left[ \begin{array}{cc|cc} 1 & 0 & \frac{1}{3} + \frac{1}{3} \cdot \frac{1}{11} & 0 + \frac{1}{3} \cdot \frac{3}{11} \\ 0 & 1 & \frac{1}{11} & \frac{3}{11} \end{array} \right] = \left[ \begin{array}{cc|cc} 1 & 0 & \frac{1}{3} + \frac{1}{33} & \frac{1}{11} \\ 0 & 1 & \frac{1}{11} & \frac{3}{11} \end{array} \right] = \left[ \begin{array}{cc|cc} 1 & 0 & \frac{11}{33} + \frac{1}{33} & \frac{1}{11} \\ 0 & 1 & \frac{1}{11} & \frac{3}{11} \end{array} \right] = \left[ \begin{array}{cc|cc} 1 & 0 & \frac{12}{33} & \frac{1}{11} \\ 0 & 1 & \frac{1}{11} & \frac{3}{11} \end{array} \right]$$ Simplify $\frac{12}{33} = \frac{4}{11}$. **Inverse of A:** $$A^{-1} = \begin{bmatrix} \frac{4}{11} & \frac{1}{11} \\ \frac{1}{11} & \frac{3}{11} \end{bmatrix}$$ 4. **Matrix B:** Start with $\left[ \begin{array}{cc|cc} -3 & 2 & 1 & 0 \\ 6 & -4 & 0 & 1 \end{array} \right]$. Step 1: Make the leading 1 in row 1 by dividing row 1 by $-3$: $$\left[ \begin{array}{cc|cc} \cancel{-3} \frac{1}{\cancel{-3}} & -\frac{2}{3} & -\frac{1}{3} & 0 \\ 6 & -4 & 0 & 1 \end{array} \right] = \left[ \begin{array}{cc|cc} 1 & -\frac{2}{3} & -\frac{1}{3} & 0 \\ 6 & -4 & 0 & 1 \end{array} \right]$$ Step 2: Eliminate the 6 in row 2, column 1 by subtracting row 1 multiplied by 6 from row 2: $$\left[ \begin{array}{cc|cc} 1 & -\frac{2}{3} & -\frac{1}{3} & 0 \\ 0 & -4 - 6 \cdot \left(-\frac{2}{3}\right) & 0 - 6 \cdot \left(-\frac{1}{3}\right) & 1 - 6 \cdot 0 \end{array} \right] = \left[ \begin{array}{cc|cc} 1 & -\frac{2}{3} & -\frac{1}{3} & 0 \\ 0 & -4 + 4 & 0 + 2 & 1 \end{array} \right] = \left[ \begin{array}{cc|cc} 1 & -\frac{2}{3} & -\frac{1}{3} & 0 \\ 0 & 0 & 2 & 1 \end{array} \right]$$ Step 3: The second row has zeros in the coefficient part but nonzero entries in the augmented part, indicating no solution to make the left side identity. **Conclusion:** Matrix B is singular and does not have an inverse. **Final answers:** $$A^{-1} = \begin{bmatrix} \frac{4}{11} & \frac{1}{11} \\ \frac{1}{11} & \frac{3}{11} \end{bmatrix}, \quad B^{-1} \text{ does not exist}.$$