solve the following linear equation using Gauss Elimination method. 2x_1+5x_2+x_3+8x_4=5, x_1+6x_2+3x_3+5x_4=11, 7x_1+2x_2+6x_3+3x_4=14, 4x_1+8x_2+x_3+2x_4=19.

To solve the system of linear equations using the **Gauss Elimination method**, we will proceed through the following steps:

Given system of equations:

\[
\begin{aligned}
1) & \ 2x_1 + 5x_2 + x_3 + 8x_4 = 5 \\
2) & \ x_1 + 6x_2 + 3x_3 + 5x_4 = 11 \\
3) & \ 7x_1 + 2x_2 + 6x_3 + 3x_4 = 14 \\
4) & \ 4x_1 + 8x_2 + x_3 + 2x_4 = 19 \\
\end{aligned}
\]

### Step 1: Form the augmented matrix

The augmented matrix for this system is:

\[
\left[
\begin{array}{cccc|c}
2 & 5 & 1 & 8 & 5 \\
1 & 6 & 3 & 5 & 11 \\
7 & 2 & 6 & 3 & 14 \\
4 & 8 & 1 & 2 & 19 \\
\end{array}
\right]
\]

### Step 2: Perform Gaussian elimination to get the matrix in row echelon form

We'll perform row operations to eliminate the lower triangular part of the matrix. Let’s proceed with the operations:

1. **R2 = R2 - (1/2) * R1** (to eliminate the \(x_1\) term in row 2)
2. **R3 = R3 - (7/2) * R1** (to eliminate the \(x_1\) term in row 3)
3. **R4 = R4 - 2 * R1** (to eliminate the \(x_1\) term in row 4)

After these operations, the matrix becomes:

\[
\left[
\begin{array}{cccc|c}
2 & 5 & 1 & 8 & 5 \\
0 & 3.5 & 2.5 & 1 & 8.5 \\
0 & -15.5 & 2.5 & -25 & -3.5 \\
0 & -2 & -1 & -14 & 9 \\
\end{array}
\right]
\]

Next, we perform the following operations to eliminate the \(x_2\) term in row 3 and row 4:

4. **R3 = R3 - (-15.5 / 3.5) * R2**
5. **R4 = R4 - (-2 / 3.5) * R2**

The resulting matrix will be in row echelon form. Let's calculate these steps now.The augmented matrix in row echelon form after performing Gaussian elimination is:

\[
\left[
\begin{array}{cccc|c}
2 & 5 & 1 & 8 & 5 \\
0 & 3.5 & 2.5 & 1 & 8.5 \\
0 & 0 & 13.57 & -20.57 & 34.14 \\
0 & 0 & 0.43 & -13.43 & 13.86 \\
\end{array}
\right]
\]

### Step 3: Perform Back Substitution

We now solve the system starting from the last row:

1. From row 4: \( 0.43x_3 - 13.43x_4 = 13.86 \)
   
   Solve for \(x_3\) and \(x_4\).

2. Use the results from \(x_3\) and \(x_4\) to substitute into row 3, then row 2, and finally row 1.

Let's calculate the back substitution steps.The solution to the system of equations is:

\[
x_1 = 1.043, \quad x_2 = 2.044, \quad x_3 = 0.952, \quad x_4 = -1.032
\]

This was obtained using the Gauss Elimination method, followed by back substitution.

Would you like further details on the steps or have any questions?

### Related Questions:
1. How does Gauss-Jordan elimination differ from Gauss elimination?
2. Can Gaussian elimination be applied to non-square matrices?
3. What are the limitations of Gaussian elimination in solving linear systems?
4. How can we check the accuracy of the solution obtained by Gaussian elimination?
5. What are the computational complexities of the Gauss elimination method?

**Tip:** Always verify your results by plugging the solution back into the original equations.

Solve the following linear equation using Gauss Elimination method. 2x_1+5x_2+x_3+8x_4=5, x_1+6x_2+3x_3+5x_4=11, 7x_1+2x_2+6x_3+3x_4=14, 4x_1+8x_2+x_3+2x_4=19.

Learn how to solve a system of linear equations using the Gauss Elimination method with a detailed, step-by-step breakdown. This problem involves eliminating variables to reduce the system into row echelon form and using back substitution to find the solution for each variable. Ideal for students studying Linear Algebra at the undergraduate level.

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