Solve the following system of linear equations by Gaussian elimination: 4x1+6x2+x3=2 (R1), 2x1+x2-4x3=3 (R2), 3x1-2x2+5x3=8 (R3)

To solve the system of linear equations using Gaussian elimination, we need to perform a series of row operations to convert the system into an upper triangular form, and then back-substitute to find the solutions.

### Step 1: Write the system as an augmented matrix
Given the system:

\[
\begin{aligned}
4x_1 + 6x_2 + x_3 &= 2 &\quad \text{(R1)} \\
2x_1 + x_2 - 4x_3 &= 3 &\quad \text{(R2)} \\
3x_1 - 2x_2 + 5x_3 &= 8 &\quad \text{(R3)}
\end{aligned}
\]

The augmented matrix is:

\[
\begin{pmatrix}
4 & 6 & 1 & \vert & 2 \\
2 & 1 & -4 & \vert & 3 \\
3 & -2 & 5 & \vert & 8
\end{pmatrix}
\]

### Step 2: Perform row operations to create zeros below the pivot in the first column
We will create zeros below the first pivot (which is 4 in R1):

- **Operation 1:** Replace \( R2 \) with \( R2 - \frac{1}{2}R1 \)
- **Operation 2:** Replace \( R3 \) with \( R3 - \frac{3}{4}R1 \)

Performing these operations:

\[
R2 \rightarrow R2 - \frac{1}{2}R1 = \left( 2 - \frac{1}{2}(4),\ 1 - \frac{1}{2}(6),\ -4 - \frac{1}{2}(1),\ 3 - \frac{1}{2}(2) \right) = (0, -2, -\frac{9}{2}, 2)
\]

\[
R3 \rightarrow R3 - \frac{3}{4}R1 = \left( 3 - \frac{3}{4}(4),\ -2 - \frac{3}{4}(6),\ 5 - \frac{3}{4}(1),\ 8 - \frac{3}{4}(2) \right) = (0, -6.5, 4.25, 6.5)
\]

The augmented matrix becomes:

\[
\begin{pmatrix}
4 & 6 & 1 & \vert & 2 \\
0 & -2 & -\frac{9}{2} & \vert & 2 \\
0 & -6.5 & 4.25 & \vert & 6.5
\end{pmatrix}
\]

### Step 3: Perform row operations to create zeros below the pivot in the second column
We will create a zero below the pivot in the second column (which is now \(-2\) in R2):

- **Operation 3:** Replace \( R3 \) with \( R3 - \frac{-6.5}{-2}R2 \)

\[
R3 \rightarrow R3 - \frac{-6.5}{-2}R2 = \left( 0,\ -6.5 - 3.25(-2),\ 4.25 - 3.25\left(-\frac{9}{2}\right),\ 6.5 - 3.25(2) \right) = (0, 0, -12.375, 0)
\]

The augmented matrix becomes:

\[
\begin{pmatrix}
4 & 6 & 1 & \vert & 2 \\
0 & -2 & -\frac{9}{2} & \vert & 2 \\
0 & 0 & -12.375 & \vert & 0
\end{pmatrix}
\]

### Step 4: Back substitution to find the solutions
Now that we have an upper triangular matrix, we can back-substitute to find the values of \(x_1\), \(x_2\), and \(x_3\).

From the third row:
\[
-12.375x_3 = 0 \quad \Rightarrow \quad x_3 = 0
\]

Substituting \(x_3 = 0\) into the second row:
\[
-2x_2 - \frac{9}{2}(0) = 2 \quad \Rightarrow \quad -2x_2 = 2 \quad \Rightarrow \quad x_2 = -1
\]

Substituting \(x_2 = -1\) and \(x_3 = 0\) into the first row:
\[
4x_1 + 6(-1) + 1(0) = 2 \quad \Rightarrow \quad 4x_1 - 6 = 2 \quad \Rightarrow \quad 4x_1 = 8 \quad \Rightarrow \quad x_1 = 2
\]

### Final Answer:
The solution to the system is:

\[
x_1 = 2,\quad x_2 = -1,\quad x_3 = 0
\]

Would you like more details or have any questions?

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Here are some related questions you might find interesting:

1. How can Gaussian elimination be used to determine if a system has no solutions?
2. What is the difference between Gaussian elimination and Gauss-Jordan elimination?
3. Can Gaussian elimination be applied to non-square matrices?
4. How does the determinant of the coefficient matrix affect the solutions of a system?
5. What are the computational complexities of Gaussian elimination?

**Tip:** Always double-check each row operation when performing Gaussian elimination to avoid calculation errors.

Learn how to solve a system of linear equations using Gaussian elimination method with step-by-step explanation. This problem covers key concepts in linear algebra and Gaussian elimination techniques. Suitable for undergraduate students and those preparing for advanced mathematics exams.

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