Math Problem Statement

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.

Solution

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

Given system of equations:

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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Linear Algebra
Systems of Linear Equations
Gaussian Elimination

Formulas

Gaussian elimination involves converting a system of linear equations into row echelon form and then solving using back substitution.

Theorems

Gauss Elimination Method
Back Substitution

Suitable Grade Level

Undergraduate Math (Linear Algebra)