1. Stating the problem: Solve the system of linear equations using the Gauss-Jordan method. Given: $$2x - 6y + 8z = 24$$ $$5x + 4y - 3z = 2$$ $$3x + y + 2z = 16$$ 2. Write the augmented matrix of the system: $$\begin{bmatrix} 2 & -6 & 8 & | & 24 \\ 5 & 4 & -3 & | & 2 \\ 3 & 1 & 2 & | & 16 \end{bmatrix}$$ 3. Make the element in row 1, column 1 a 1 by dividing the first row by 2: $$R_1 \to \frac{1}{2}R_1 = \begin{bmatrix}1 & -3 & 4 & | & 12\end{bmatrix}$$ Now matrix is: $$\begin{bmatrix} 1 & -3 & 4 & | & 12 \\ 5 & 4 & -3 & | & 2 \\ 3 & 1 & 2 & | & 16 \end{bmatrix}$$ 4. Eliminate the x terms in rows 2 and 3: - Row 2: $R_2 = R_2 - 5R_1 = \begin{bmatrix}5 - 5(1) & 4 - 5(-3) & -3 - 5(4) & 2 - 5(12)\end{bmatrix} = \begin{bmatrix}0 & 19 & -23 & -58\end{bmatrix}$ - Row 3: $R_3 = R_3 - 3R_1 = \begin{bmatrix}3 - 3(1) & 1 - 3(-3) & 2 - 3(4) & 16 - 3(12)\end{bmatrix} = \begin{bmatrix}0 & 10 & -10 & -20\end{bmatrix}$ Matrix is now: $$\begin{bmatrix} 1 & -3 & 4 & | & 12 \\ 0 & 19 & -23 & | & -58 \\ 0 & 10 & -10 & | & -20 \end{bmatrix}$$ 5. Make the leading coefficient in row 2 a 1 by dividing row 2 by 19: $$R_2 \to \frac{1}{19} R_2 = \begin{bmatrix}0 & 1 & -\frac{23}{19} & | & -\frac{58}{19}\end{bmatrix}$$ Matrix: $$\begin{bmatrix} 1 & -3 & 4 & | & 12 \\ 0 & 1 & -\frac{23}{19} & | & -\frac{58}{19} \\ 0 & 10 & -10 & | & -20 \end{bmatrix}$$ 6. Use row 2 to eliminate the y term in rows 1 and 3: - Row 1: $R_1 = R_1 + 3R_2 = \begin{bmatrix}1 & -3 + 3(1) & 4 + 3(-\frac{23}{19}) & 12 + 3(-\frac{58}{19})\end{bmatrix} = \begin{bmatrix}1 & 0 & \frac{14}{19} & \frac{126}{19}\end{bmatrix}$ - Row 3: $R_3 = R_3 - 10R_2 = \begin{bmatrix}0 & 10 - 10(1) & -10 - 10(-\frac{23}{19}) & -20 - 10(-\frac{58}{19})\end{bmatrix} = \begin{bmatrix}0 & 0 & \frac{130}{19} & \frac{260}{19}\end{bmatrix}$ Matrix is now: $$\begin{bmatrix} 1 & 0 & \frac{14}{19} & | & \frac{126}{19} \\ 0 & 1 & -\frac{23}{19} & | & -\frac{58}{19} \\ 0 & 0 & \frac{130}{19} & | & \frac{260}{19} \end{bmatrix}$$ 7. Make the leading coefficient in row 3 a 1 by multiplying row 3 by $\frac{19}{130}$: $$R_3 \to \frac{19}{130} R_3 = \begin{bmatrix}0 & 0 & 1 & | & 2\end{bmatrix}$$ Matrix is: $$\begin{bmatrix} 1 & 0 & \frac{14}{19} & | & \frac{126}{19} \\ 0 & 1 & -\frac{23}{19} & | & -\frac{58}{19} \\ 0 & 0 & 1 & | & 2 \end{bmatrix}$$ 8. Eliminate the z terms from rows 1 and 2: - Row 1: $R_1 = R_1 - \frac{14}{19} R_3 = \begin{bmatrix}1 & 0 & 0 & | & \frac{126}{19} - \frac{14}{19} \times 2\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 & | & 6\end{bmatrix}$ - Row 2: $R_2 = R_2 + \frac{23}{19} R_3 = \begin{bmatrix}0 & 1 & 0 & | & -\frac{58}{19} + \frac{23}{19} \times 2\end{bmatrix} = \begin{bmatrix}0 & 1 & 0 & | & 2\end{bmatrix}$ Final matrix: $$\begin{bmatrix} 1 & 0 & 0 & | & 6 \\ 0 & 1 & 0 & | & 2 \\ 0 & 0 & 1 & | & 2 \end{bmatrix}$$ 9. Read off solutions: $$x = 6, \quad y = 2, \quad z = 2$$ Answer: $\boxed{x=6, y=2, z=2}$