1. **State the problem:** Find the inverse of matrix $$A = \begin{pmatrix} 2 & 4 & 2 \\ -3 & 4 & 3 \\ -1 & 1 & 3 \end{pmatrix}$$ using the Gauss-Jordan elimination method. 2. **Method overview:** To find the inverse of a matrix using Gauss-Jordan elimination, we augment matrix $$A$$ with the identity matrix $$I$$ of the same size and perform row operations to transform $$A$$ into $$I$$. The augmented part will then become $$A^{-1}$$. 3. **Set up the augmented matrix:** $$\left( \begin{array}{ccc|ccc} 2 & 4 & 2 & 1 & 0 & 0 \\ -3 & 4 & 3 & 0 & 1 & 0 \\ -1 & 1 & 3 & 0 & 0 & 1 \end{array} \right)$$ 4. **Step 1: Make the pivot in row 1, column 1 equal to 1 by dividing row 1 by 2:** $$\left( \begin{array}{ccc|ccc} \cancel{2} & 4 & 2 & 1 & 0 & 0 \\ -3 & 4 & 3 & 0 & 1 & 0 \\ -1 & 1 & 3 & 0 & 0 & 1 \end{array} \right) \to \left( \begin{array}{ccc|ccc} 1 & 2 & 1 & \frac{1}{2} & 0 & 0 \\ -3 & 4 & 3 & 0 & 1 & 0 \\ -1 & 1 & 3 & 0 & 0 & 1 \end{array} \right)$$ 5. **Step 2: Eliminate entries below and above the pivot in column 1:** - Add 3 times row 1 to row 2: $$R_2 = R_2 + 3R_1$$ - Add 1 times row 1 to row 3: $$R_3 = R_3 + R_1$$ Resulting matrix: $$\left( \begin{array}{ccc|ccc} 1 & 2 & 1 & \frac{1}{2} & 0 & 0 \\ 0 & 10 & 6 & \frac{3}{2} & 1 & 0 \\ 0 & 3 & 4 & \frac{1}{2} & 0 & 1 \end{array} \right)$$ 6. **Step 3: Make the pivot in row 2, column 2 equal to 1 by dividing row 2 by 10:** $$\left( \begin{array}{ccc|ccc} 1 & 2 & 1 & \frac{1}{2} & 0 & 0 \\ 0 & \cancel{10} & 6 & \frac{3}{2} & 1 & 0 \\ 0 & 3 & 4 & \frac{1}{2} & 0 & 1 \end{array} \right) \to \left( \begin{array}{ccc|ccc} 1 & 2 & 1 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & \frac{6}{10} & \frac{3}{20} & \frac{1}{10} & 0 \\ 0 & 3 & 4 & \frac{1}{2} & 0 & 1 \end{array} \right)$$ 7. **Step 4: Eliminate entries above and below the pivot in column 2:** - Subtract 2 times row 2 from row 1: $$R_1 = R_1 - 2R_2$$ - Subtract 3 times row 2 from row 3: $$R_3 = R_3 - 3R_2$$ Resulting matrix: $$\left( \begin{array}{ccc|ccc} 1 & 0 & 1 - 2 \times \frac{6}{10} & \frac{1}{2} - 2 \times \frac{3}{20} & 0 - 2 \times \frac{1}{10} & 0 \\ 0 & 1 & \frac{6}{10} & \frac{3}{20} & \frac{1}{10} & 0 \\ 0 & 0 & 4 - 3 \times \frac{6}{10} & \frac{1}{2} - 3 \times \frac{3}{20} & 0 - 3 \times \frac{1}{10} & 1 \end{array} \right)$$ Simplify: - Row 1, column 3: $1 - 2 \times \frac{6}{10} = 1 - \frac{12}{10} = 1 - 1.2 = -0.2 = -\frac{1}{5}$ - Row 1, augmented part: - $\frac{1}{2} - 2 \times \frac{3}{20} = \frac{1}{2} - \frac{6}{20} = \frac{1}{2} - \frac{3}{10} = \frac{5}{10} - \frac{3}{10} = \frac{2}{10} = \frac{1}{5}$ - $0 - 2 \times \frac{1}{10} = 0 - \frac{2}{10} = -\frac{1}{5}$ - $0$ - Row 3, column 3: $4 - 3 \times \frac{6}{10} = 4 - \frac{18}{10} = 4 - 1.8 = 2.2 = \frac{11}{5}$ - Row 3, augmented part: - $\frac{1}{2} - 3 \times \frac{3}{20} = \frac{1}{2} - \frac{9}{20} = \frac{10}{20} - \frac{9}{20} = \frac{1}{20}$ - $0 - 3 \times \frac{1}{10} = 0 - \frac{3}{10} = -\frac{3}{10}$ - $1$ So matrix becomes: $$\left( \begin{array}{ccc|ccc} 1 & 0 & -\frac{1}{5} & \frac{1}{5} & -\frac{1}{5} & 0 \\ 0 & 1 & \frac{3}{5} & \frac{3}{20} & \frac{1}{10} & 0 \\ 0 & 0 & \frac{11}{5} & \frac{1}{20} & -\frac{3}{10} & 1 \end{array} \right)$$ 8. **Step 5: Make the pivot in row 3, column 3 equal to 1 by dividing row 3 by $\frac{11}{5}$:** $$\left( \begin{array}{ccc|ccc} 1 & 0 & -\frac{1}{5} & \frac{1}{5} & -\frac{1}{5} & 0 \\ 0 & 1 & \frac{3}{5} & \frac{3}{20} & \frac{1}{10} & 0 \\ 0 & 0 & \cancel{\frac{11}{5}} & \frac{1}{20} & -\frac{3}{10} & 1 \end{array} \right) \to \left( \begin{array}{ccc|ccc} 1 & 0 & -\frac{1}{5} & \frac{1}{5} & -\frac{1}{5} & 0 \\ 0 & 1 & \frac{3}{5} & \frac{3}{20} & \frac{1}{10} & 0 \\ 0 & 0 & 1 & \frac{1}{20} \times \frac{5}{11} & -\frac{3}{10} \times \frac{5}{11} & 1 \times \frac{5}{11} \end{array} \right)$$ Simplify row 3 augmented part: - $\frac{1}{20} \times \frac{5}{11} = \frac{5}{220} = \frac{1}{44}$ - $-\frac{3}{10} \times \frac{5}{11} = -\frac{15}{110} = -\frac{3}{22}$ - $1 \times \frac{5}{11} = \frac{5}{11}$ Matrix now: $$\left( \begin{array}{ccc|ccc} 1 & 0 & -\frac{1}{5} & \frac{1}{5} & -\frac{1}{5} & 0 \\ 0 & 1 & \frac{3}{5} & \frac{3}{20} & \frac{1}{10} & 0 \\ 0 & 0 & 1 & \frac{1}{44} & -\frac{3}{22} & \frac{5}{11} \end{array} \right)$$ 9. **Step 6: Eliminate entries above the pivot in column 3:** - Add $\frac{1}{5}$ times row 3 to row 1: $$R_1 = R_1 + \frac{1}{5} R_3$$ - Subtract $\frac{3}{5}$ times row 3 from row 2: $$R_2 = R_2 - \frac{3}{5} R_3$$ Calculate row 1 augmented part: - $\frac{1}{5} + \frac{1}{5} \times \frac{1}{44} = \frac{1}{5} + \frac{1}{220} = \frac{44}{220} + \frac{1}{220} = \frac{45}{220} = \frac{9}{44}$ - $-\frac{1}{5} + \frac{1}{5} \times -\frac{3}{22} = -\frac{1}{5} - \frac{3}{110} = -\frac{22}{110} - \frac{3}{110} = -\frac{25}{110} = -\frac{5}{22}$ - $0 + \frac{1}{5} \times \frac{5}{11} = 0 + \frac{1}{11} = \frac{1}{11}$ Calculate row 2 augmented part: - $\frac{3}{20} - \frac{3}{5} \times \frac{1}{44} = \frac{3}{20} - \frac{3}{220} = \frac{33}{220} - \frac{3}{220} = \frac{30}{220} = \frac{3}{22}$ - $\frac{1}{10} - \frac{3}{5} \times -\frac{3}{22} = \frac{1}{10} + \frac{9}{110} = \frac{11}{110} + \frac{9}{110} = \frac{20}{110} = \frac{2}{11}$ - $0 - \frac{3}{5} \times \frac{5}{11} = 0 - \frac{15}{55} = -\frac{3}{11}$ Matrix now: $$\left( \begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{9}{44} & -\frac{5}{22} & \frac{1}{11} \\ 0 & 1 & 0 & \frac{3}{22} & \frac{2}{11} & -\frac{3}{11} \\ 0 & 0 & 1 & \frac{1}{44} & -\frac{3}{22} & \frac{5}{11} \end{array} \right)$$ 10. **Final result:** The right side of the augmented matrix is the inverse: $$A^{-1} = \begin{pmatrix} \frac{9}{44} & -\frac{5}{22} & \frac{1}{11} \\ \frac{3}{22} & \frac{2}{11} & -\frac{3}{11} \\ \frac{1}{44} & -\frac{3}{22} & \frac{5}{11} \end{pmatrix}$$ This completes the Gauss-Jordan elimination to find the inverse of matrix $$A$$.