Sure! Gaussian elimination is a method used to solve systems of linear equations. It involves performing row operations to reduce a matrix into an upper triangular form and then using back substitution to find the solution. Here, I'll provide a worked example of solving a system of linear equations using Gaussian elimination.
Example Problem
Let's solve the following system of linear equations:
- \( 2x + 3y + z = 1 \)
- \( 4x + y - 2z = -2 \)
- \( -2x + y + 3z = 3 \)
Step 1: Write the Augmented Matrix
We start by writing the augmented matrix corresponding to the system of equations:
\[ \begin{bmatrix} 2 & 3 & 1 & | & 1 \ 4 & 1 & -2 & | & -2 \ -2 & 1 & 3 & | & 3 \end{bmatrix} \]
Step 2: Apply Gaussian Elimination
Row operation to obtain zeros below the pivot element in Column 1:
- \( R_2 \rightarrow R_2 - 2R_1 \)
- \( R_3 \rightarrow R_3 + R_1 \)
Calculating this, we have:
- For \( R_2 \): \( 4 - 2(2) = 0, 1 - 2(3) = -5, -2 - 2(1) = -4, -2 - 2(1) = -3 \)
- For \( R_3 \): \( -2 + 2 = 0, 1 + 3 = 4, 3 + 1 = 4, 3 + 1 = 4 \)
The matrix now looks like:
\[ \begin{bmatrix} 2 & 3 & 1 & | & 1 \ 0 & -5 & -4 & | & -3 \ 0 & 4 & 4 & | & 4 \end{bmatrix} \]
Next, we eliminate the second row to obtain zeros below \( R_2[2, 2] \):
- \( R_3 \rightarrow R_3 + \frac{4}{5}R_2 \)
Calculating this:
- For \( R_3 \): \( 0 \cdot 0 + \frac{4}{5}(0) = 0 \) \( 4 + \frac{4}{5}(-5) = 0 \) \( 4 + \frac{4}{5}(-4) = \frac{16}{5} \) \( 4 + (-\frac{12}{5}) = \frac{8}{5} \)
Thus, the new augmented matrix becomes:
\[ \begin{bmatrix} 2 & 3 & 1 & | & 1 \ 0 & -5 & -4 & | & -3 \ 0 & 0 & \frac{16}{5} & | & \frac{8}{5} \end{bmatrix} \]
Step 3: Back Substitution
Now we can perform back substitution:
From Row 3, we have: \[ \frac{16}{5} z = \frac{8}{5} \implies z = \frac{8/5}{16/5} = \frac{1}{2} \]
Now substitute \( z \) into Row 2: \[ -5y - 4 \left( \frac{1}{2} \right) = -3 \ -5y - 2 = -3 \ -5y = -1 \implies y = \frac{1}{5} \]
Finally, substitute \( y \) and \( z \) back into Row 1: \[ 2x + 3\left( \frac{1}{5} \right) + \frac{1}{2} = 1 \ 2x + \frac{3}{5} + \frac{5}{10} = 1 \ 2x + \frac{3}{5} + \frac{1}{2} = 1 \] Converting \( \frac{3}{5} = \frac{6}{10} \) gives: \[ 2x + \frac{6}{10} + \frac{5}{10} = 1 \ 2x + \frac{11}{10} = 1 \implies 2x = 1 - \frac{11}{10} \implies 2x = -\frac{1}{10} \implies x = -\frac{1}{20} \]
Final Solution
Putting it all together, the solution to the system of equations is: \[ x = -\frac{1}{20}, \quad y = \frac{1}{5}, \quad z = \frac{1}{2} \]
This completes the example of using Gaussian elimination to solve a system of linear equations. If you have further questions or need clarification on any steps, feel free to ask!