> [!summary] Gaussian elimination is a technique for solving a matrix into row echelon form >[!info]+ Read Time **⏱ 2 mins** # Definition Gaussian elimination is a general algorithm to solve a [[Matrix Notation|matrix]] to [[Row Echelon Form|row echelon form]]. This algorithm does not always need to be followed when solving a matrix, but in general is the fastest way to solve a matrix to get row echelon form. The steps are as follows 1. Locate the leftmost column that is not entirely of zeros 2. Interchange rows if necessary to bring a nonzero entry to the top of the column in step 1 3. If the entry is now at the top of the column found in step 1 is $a$ multiply the row by $\frac{1}{a}$ to introduce a [[Leading 1s|leading 1]] 4. Zero out all the other entries below the pivot in the same column 5. Cover the top row in the matrix and begin again at step 1 applied to the [[Sub-Matrix|submatrix]] that remains. Continue this process until the requirements for [[Row Echelon Form|row echelon form]] is made. After a matrix is in row echelon form the matrix can be converted back into [[Linear Equations|linear equations]] where each variable is solved for. # Examples >[!example] Solve the following matrix using Gaussian elimination > $ \begin{array}{c} \textbf{Given system:} \\ x + y + z = 6 \\ 2x + 3y + z = 14 \\ x + 2y + 3z = 14 \\ \\ \textbf{Step 1: Locate leftmost nonzero column (Column 1)} \\ \\ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 2 & 3 & 1 & 14 \\ 1 & 2 & 3 & 14 \end{array} \right] \\ \\ \textbf{Step 2: Interchange rows if needed (not needed)} \\ \\ \textbf{Step 3: Leading 1 already present at } a_{11} \\ \\ \textbf{Step 4: Zero out entries below the pivot} \\ R_2 \rightarrow R_2 - 2R_1, \quad R_3 \rightarrow R_3 - R_1 \\ \\ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & -1 & 2 \\ 0 & 1 & 2 & 8 \end{array} \right] \\ \\ \textbf{Step 5: Submatrix begins at Row 2. Eliminate below new pivot in Column 2} \\ R_3 \rightarrow R_3 - R_2 \\ \\ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 3 & 6 \end{array} \right] \\ \\ \textbf{Back-substitute to solve linear equations:} \\ \\ \text{From Row 3: } 3z = 6 \Rightarrow z = 2 \\ \text{From Row 2: } y - z = 2 \Rightarrow y = 2 + z = 4 \\ \text{From Row 1: } x + y + z = 6 \Rightarrow x = 6 - y - z = 0 \\ \\ \textbf{Final Answer:} \quad x = 0, \quad y = 4, \quad z = 2 \end{array} > $ # Resources <iframe width="560" height="315" src="https://www.youtube.com/embed/seet9VyHo3Q?si=-54HI5yVDo6MnTOO" title="YouTube video player" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" referrerpolicy="strict-origin-when-cross-origin" allowfullscreen></iframe>