Question

Worked examples using Gaussian elimination

Answer 1

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:

1. \( 2x + 3y + z = 1 \)
2. \( 4x + y - 2z = -2 \)
3. \( -2x + y + 3z = 3 \)

### 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:**
1. \( R_2 \rightarrow R_2 - 2R_1 \)
2. \( R_3 \rightarrow R_3 + R_1 \)

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 \( R_2[2, 2] \):**
1. \( R_3 \rightarrow R_3 + \frac{4}{5}R_2 \)

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!