1. **Stating the problem:** We have the system of linear equations: $$\begin{cases} x_1 + x_3 = 3 \\ x_1 - x_2 - x_3 = 1 \\ - x_1 + x_2 = 4 \end{cases}$$ We want to find which matrix among the options A, B, C, and D is the upper triangular coefficient matrix of this system. 2. **Recall:** An upper triangular matrix has all zeros below the main diagonal. The coefficient matrix of the system is: $$\begin{bmatrix} 1 & 0 & 1 \\ 1 & -1 & -1 \\ -1 & 1 & 0 \end{bmatrix}$$ (from the coefficients of $x_1, x_2, x_3$ respectively). 3. **Goal:** Use Gaussian elimination to convert the coefficient matrix into an upper triangular matrix. 4. **Step 1:** Start with the matrix: $$A = \begin{bmatrix} 1 & 0 & 1 \\ 1 & -1 & -1 \\ -1 & 1 & 0 \end{bmatrix}$$ 5. **Step 2:** Eliminate the $x_1$ terms below the first row. - Row 2: $R_2 - R_1 \to R_2$ $$\begin{bmatrix} 1 & 0 & 1 \\ 1-1 & -1-0 & -1-1 \\ -1 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & -1 & -2 \\ -1 & 1 & 0 \end{bmatrix}$$ - Row 3: $R_3 + R_1 \to R_3$ $$\begin{bmatrix} 1 & 0 & 1 \\ 0 & -1 & -2 \\ -1+1 & 1+0 & 0+1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & -1 & -2 \\ 0 & 1 & 1 \end{bmatrix}$$ 6. **Step 3:** Eliminate the $x_2$ term below the second row. - Row 3: $R_3 + R_2 \to R_3$ $$\begin{bmatrix} 1 & 0 & 1 \\ 0 & -1 & -2 \\ 0 & 1-1 & 1-2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & -1 & -2 \\ 0 & 0 & -1 \end{bmatrix}$$ 7. **Result:** The upper triangular matrix is $$\begin{bmatrix} 1 & 0 & 1 \\ 0 & -1 & -2 \\ 0 & 0 & -1 \end{bmatrix}$$ 8. **Compare with options:** This matches option C. **Final answer:** Option C is the upper triangular coefficient matrix of the system.