Math Problem Statement

Solve the system of equations using matrices. Use the Gaussian elimination method with​ back-substitution. left brace Start 3 By 1 Matrix 1st Row 1st Column nothing 2nd Row 1st Column nothing 3rd Row 1st Column nothing EndMatrix

w minus x minus y minus z equals 8 4 w minus 3 x minus y plus z equals 7 3 w plus x minus 3 y minus z equals 26 negative w plus 3 x plus 2 y plus 3 z equals negative 13 Question content area bottom Part 1 The solution set is StartSet left parenthesis nothing comma nothing comma nothing comma nothing right parenthesis EndSet             . ​(Simplify your​ answers.)

Solution

To solve the system of equations using Gaussian elimination and back-substitution, we first convert the system of equations into an augmented matrix and then proceed with row operations to bring the matrix to row-echelon form. Afterward, we use back-substitution to find the values of w,x,y,w, x, y, and zz.

Step 1: Write the system of equations

The system of equations is:

1. & \quad w - x - y - z = 8 \\ 2. & \quad 4w - 3x - y + z = 7 \\ 3. & \quad 3w + x - 3y - z = 26 \\ 4. & \quad -w + 3x + 2y + 3z = -13 \end{aligned}$$ ### Step 2: Set up the augmented matrix We can write the system of equations as the following augmented matrix: $$\left[ \begin{array}{cccc|c} 1 & -1 & -1 & -1 & 8 \\ 4 & -3 & -1 & 1 & 7 \\ 3 & 1 & -3 & -1 & 26 \\ -1 & 3 & 2 & 3 & -13 \end{array} \right]$$ ### Step 3: Perform Gaussian elimination We will now use row operations to bring the matrix into upper triangular (row-echelon) form. #### First pivot (make the first column 1): The first element of the first row is already 1. So, no changes are needed for the first row. #### Eliminate the entries below the first pivot: We use row operations to eliminate the first entries in rows 2, 3, and 4. - $$R_2 \rightarrow R_2 - 4R_1$$ - $$R_3 \rightarrow R_3 - 3R_1$$ - $$R_4 \rightarrow R_4 + R_1$$ After these operations, the matrix becomes: $$\left[ \begin{array}{cccc|c} 1 & -1 & -1 & -1 & 8 \\ 0 & 1 & 3 & 5 & -25 \\ 0 & 4 & 0 & 2 & 2 \\ 0 & 2 & 1 & 2 & -5 \end{array} \right]$$ #### Second pivot (make the second column 1): We divide row 2 by 1 (it’s already 1, so no change is needed). #### Eliminate the entries below the second pivot: We eliminate the second entries in rows 3 and 4 using row operations: - $$R_3 \rightarrow R_3 - 4R_2$$ - \( R_4 \rightarrow

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 Equations
Gaussian Elimination
Back-Substitution

Formulas

Row operations for Gaussian elimination
Back-substitution for solving upper triangular matrices

Theorems

Gaussian elimination

Suitable Grade Level

Grades 10-12

Related Recommendation

Solve System of Equations Using Gaussian Elimination: Example with Four Variables

Solving a 3-Variable System of Linear Equations using Matrices: Gaussian Elimination

Solving a System of Linear Equations Using Gaussian Elimination

Solving Systems of Linear Equations Using Gaussian Elimination or Gauss-Jordan Elimination

Solving a System of Linear Equations using Gaussian Elimination