1. **State the problem:** Solve the system of linear equations using row operations and find the solutions. 2. **Initial matrix:** $$\begin{bmatrix} 2 & -3 & -4 & 26 \\ -4 & 5 & -2 & 5 \\ -8 & 10 & -6 & 12 \end{bmatrix}$$ 3. **Perform the row operation** $\frac{1}{2}R_1 \to R_1$: $$\begin{bmatrix} \cancel{2} \times \frac{1}{2} & -3 \times \frac{1}{2} & -4 \times \frac{1}{2} & 26 \times \frac{1}{2} \\ -4 & 5 & -2 & 5 \\ -8 & 10 & -6 & 12 \end{bmatrix} = \begin{bmatrix} 1 & -\frac{3}{2} & -2 & 13 \\ -4 & 5 & -2 & 5 \\ -8 & 10 & -6 & 12 \end{bmatrix}$$ 4. **Perform the operations** $+4R_1 + R_2 \to R_2$ and $+8R_1 + R_3 \to R_3$: For $R_2$: $$4 \times \begin{bmatrix} 1 & -\frac{3}{2} & -2 & 13 \end{bmatrix} + \begin{bmatrix} -4 & 5 & -2 & 5 \end{bmatrix} = \begin{bmatrix} 4 & -6 & -8 & 52 \end{bmatrix} + \begin{bmatrix} -4 & 5 & -2 & 5 \end{bmatrix} = \begin{bmatrix} 0 & -1 & -10 & 57 \end{bmatrix}$$ For $R_3$: $$8 \times \begin{bmatrix} 1 & -\frac{3}{2} & -2 & 13 \end{bmatrix} + \begin{bmatrix} -8 & 10 & -6 & 12 \end{bmatrix} = \begin{bmatrix} 8 & -12 & -16 & 104 \end{bmatrix} + \begin{bmatrix} -8 & 10 & -6 & 12 \end{bmatrix} = \begin{bmatrix} 0 & -2 & -22 & 116 \end{bmatrix}$$ Resulting matrix: $$\begin{bmatrix} 1 & -\frac{3}{2} & -2 & 13 \\ 0 & -1 & -10 & 57 \\ 0 & -2 & -22 & 116 \end{bmatrix}$$ 5. **Finish simplifying to reduced row echelon form:** Divide $R_2$ by $-1$: $$\begin{bmatrix} 1 & -\frac{3}{2} & -2 & 13 \\ 0 & \cancel{-1} \times \frac{1}{-1} & 10 \times \frac{1}{-1} & 57 \times \frac{1}{-1} \\ 0 & -2 & -22 & 116 \end{bmatrix} = \begin{bmatrix} 1 & -\frac{3}{2} & -2 & 13 \\ 0 & 1 & 10 & -57 \\ 0 & -2 & -22 & 116 \end{bmatrix}$$ Add $2R_2$ to $R_3$: $$\begin{bmatrix} 0 & -2 & -22 & 116 \end{bmatrix} + 2 \times \begin{bmatrix} 0 & 1 & 10 & -57 \end{bmatrix} = \begin{bmatrix} 0 & 0 & -2 & 2 \end{bmatrix}$$ Divide $R_3$ by $-2$: $$\begin{bmatrix} 0 & 0 & \cancel{-2} \times \frac{1}{-2} & 2 \times \frac{1}{-2} \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 & -1 \end{bmatrix}$$ Eliminate $x_3$ from $R_1$ and $R_2$: $R_1 + 2R_3 \to R_1$: $$\begin{bmatrix} 1 & -\frac{3}{2} & -2 & 13 \end{bmatrix} + 2 \times \begin{bmatrix} 0 & 0 & 1 & -1 \end{bmatrix} = \begin{bmatrix} 1 & -\frac{3}{2} & 0 & 11 \end{bmatrix}$$ $R_2 - 10R_3 \to R_2$: $$\begin{bmatrix} 0 & 1 & 10 & -57 \end{bmatrix} - 10 \times \begin{bmatrix} 0 & 0 & 1 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & -47 \end{bmatrix}$$ Eliminate $x_2$ from $R_1$: $R_1 + \frac{3}{2} R_2 \to R_1$: $$\begin{bmatrix} 1 & -\frac{3}{2} & 0 & 11 \end{bmatrix} + \frac{3}{2} \times \begin{bmatrix} 0 & 1 & 0 & -47 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 1 \end{bmatrix}$$ Final reduced matrix: $$\begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -47 \\ 0 & 0 & 1 & -1 \end{bmatrix}$$ 6. **Number of solutions:** Since the matrix is in reduced row echelon form with pivots in every variable column, there is a unique solution. 7. **Solutions:** $$x_1 = 1, \quad x_2 = -47, \quad x_3 = -1$$