How to solve a system of linear equations using Gaussian elimination?

Here is the complete solution. **Allowed Row Operations:** 1. **Swap** two rows ($R_i \leftrightarrow R_j$) 2. **Scale** a row by a non-zero constant ($cR_i \ o R_i$) 3. **Add** a multiple of one row to another ($R_i + cR_j \ o R_i$) **Augmented Matrix:** $$\begin{pmatrix} 1 & 2 & 3 & | & 9 \\ 2 & -1 & 1 & | & 8 \\ 3 & 0 & -1 & | & 3 \end{pmatrix}$$ **Step 1:** Eliminate $x$ from rows 2 and 3. $R_2 \ ightarrow R_2 - 2R_1$: $$\begin{pmatrix} 1 & 2 & 3 & | & 9 \\ 0 & -5 & -5 & | & -10 \\ 3 & 0 & -1 & | & 3 \end{pmatrix}$$ $R_3 \ ightarrow R_3 - 3R_1$: $$\begin{pmatrix} 1 & 2 & 3 & | & 9 \\ 0 & -5 & -5 & | & -10 \\ 0 & -6 & -10 & | & -24 \end{pmatrix}$$ **Step 2:** Eliminate $y$ from row 3. $R_3 \ ightarrow 5R_3 - 6R_2$: $$\begin{pmatrix} 1 & 2 & 3 & | & 9 \\ 0 & -5 & -5 & | & -10 \\ 0 & 0 & -20 & | & -60 \end{pmatrix}$$ **Step 3:** Back-substitution. From $R_3$: $-20z = -60 \Rightarrow z = 3$ From $R_2$: $-5y - 5(3) = -10 \Rightarrow -5y = 5 \Rightarrow y = -1$ From $R_1$: $x + 2(-1) + 3(3) = 9 \Rightarrow x - 2 + 9 = 9 \Rightarrow x = 2$ **Answer:** $(x, y, z) = (2, -1, 3)$ **Row-Echelon vs. Reduced Row-Echelon (RREF):** | Form | Requirements | When to Use | |---|---|---| | **Row-Echelon** | Leading entry in each row is 1 (pivot), zeros below each pivot | Back-substitution | | **RREF** | Row-echelon + zeros above each pivot as well | Directly reading solutions | Gaussian elimination stops at row-echelon form, then uses back-substitution. Gauss-Jordan elimination continues to RREF, which gives the solution directly without back-substitution.