1. **Stating the problem:** We are given a system of 3 equations with 3 variables represented by an augmented matrix. The matrix undergoes a series of row operations to find the solution. 2. **Method used:** The method applied here is **Gaussian elimination** (or row reduction) to transform the matrix into reduced row echelon form (RREF). This involves using elementary row operations to simplify the system step-by-step. 3. **Explanation of the method:** - We perform row operations such as replacing a row by itself minus a multiple of another row. - The goal is to get leading 1s (pivots) in each row and zeros below and above these pivots. - Once in RREF, the solution can be read directly from the matrix. 4. **Intermediate steps:** - Initial matrix: $$\begin{bmatrix}1 & 1 & 1 & 5 \\ 2 & 3 & 5 & 8 \\ 4 & 0 & 5 & 2\end{bmatrix}$$ - After $R_2 - 2R_1 \to R_2$: $$\begin{bmatrix}1 & 1 & 1 & 5 \\ 0 & 1 & 3 & -2 \\ 4 & 0 & 5 & 2\end{bmatrix}$$ - After $R_3 - 4R_1 \to R_3$: $$\begin{bmatrix}1 & 1 & 1 & 5 \\ 0 & 1 & 3 & -2 \\ 0 & -4 & 1 & -18\end{bmatrix}$$ - After $R_3 + 4R_2 \to R_3$: $$\begin{bmatrix}1 & 1 & 1 & 5 \\ 0 & 1 & 3 & -2 \\ 0 & 0 & 13 & -26\end{bmatrix}$$ - After scaling $R_3$ by $\frac{1}{13}$: $$\begin{bmatrix}1 & 1 & 1 & 5 \\ 0 & 1 & 3 & -2 \\ 0 & 0 & 1 & -2\end{bmatrix}$$ - After $R_2 - 3R_3 \to R_2$: $$\begin{bmatrix}1 & 1 & 1 & 5 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & -2\end{bmatrix}$$ - After $R_1 - R_3 \to R_1$: $$\begin{bmatrix}1 & 1 & 0 & 7 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & -2\end{bmatrix}$$ - After $R_1 - R_2 \to R_1$: $$\begin{bmatrix}1 & 0 & 0 & 3 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & -2\end{bmatrix}$$ 5. **Final solution:** From the last matrix, the variables correspond to: $$x = 3, \quad y = 4, \quad z = -2$$ This means the solution to the system is $\boxed{(3, 4, -2)}$.