Math Problem Statement

Solve the system using either Gaussian elimination with back-substitution or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTION. If there are an infinite number of solutions, set 

x3 = t

 and solve for x1 and x2.)

x1

 − 

3x3

 = 

−5

3x1

 + 

x2

 − 

2x3

 = 

1

2x1

 + 

2x2

 + 

x3

 = 

1

(x1, x2, x3) =      

Solution

To solve the system, we can use Gaussian elimination with back-substitution. Let’s rewrite the system in matrix form:

x_1 - 3x_3 = -5 \\ 3x_1 + x_2 - 2x_3 = 1 \\ 2x_1 + 2x_2 + x_3 = 1 \end{cases}$$ ### Step 1: Write the Augmented Matrix The augmented matrix for this system is: $$\begin{bmatrix} 1 & 0 & -3 & | & -5 \\ 3 & 1 & -2 & | & 1 \\ 2 & 2 & 1 & | & 1 \\ \end{bmatrix}$$ ### Step 2: Make the First Column into Echelon Form 1. Use row 1 as the pivot. Eliminate the $$3x_1$$ in row 2 and the $$2x_1$$ in row 3. - Row 2: $$R_2 - 3 \times R_1$$: $$\begin{bmatrix} 0 & 1 & 7 & | & 16 \\ \end{bmatrix}$$ - Row

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Systems of Linear Equations
Gaussian Elimination
Gauss-Jordan Elimination

Formulas

Gaussian Elimination steps for solving systems of equations

Theorems

Gaussian Elimination Method
Gauss-Jordan Elimination Method

Suitable Grade Level

Grades 10-12