Math Problem Statement

Use the​ Gauss-Jordan method to solve the system. 2xplus 2yminus 2 z equals 22 3xminus yplus zequals negative 19 xminus 4yplus 2zequals negative 74

Solution

To solve the system of equations using the Gauss-Jordan method, we first write the augmented matrix for the system of equations:

2x + 2y - 2z &= 22 \\ 3x - y + z &= -19 \\ x - 4y + 2z &= -74 \end{align*}$$ The augmented matrix is: $$\begin{pmatrix} 2 & 2 & -2 & | & 22 \\ 3 & -1 & 1 & | & -19 \\ 1 & -4 & 2 & | & -74 \end{pmatrix}$$ **Step 1: Make the element in the first row, first column a 1.** Divide the first row by 2: $$\begin{pmatrix} 1 & 1 & -1 & | & 11 \\ 3 & -1 & 1 & | & -19 \\ 1 & -4 & 2 & | & -74 \end{pmatrix}$$ **Step 2: Make the elements below the leading 1 in the first column zero.** Subtract 3 times the first row from the second row: $$\text{Row 2} = \text{Row 2} - 3 \cdot \text{Row 1}$$ $$\begin{pmatrix} 1 & 1 & -1 & | & 11 \\ 0 & -4 & 4 & | & -52 \\ 1 & -4 & 2 & | & -74 \end{pmatrix}$$ Subtract the first row from the third row: $$\text{Row 3} = \text{Row 3} - \text{Row 1}$$ $$\begin{pmatrix} 1 & 1 & -1 & | & 11 \\ 0 & -4 & 4 & | & -52 \\ 0 & -5 & 3 & | & -85 \end{pmatrix}$$ **Step 3: Make the element in the second row, second column a 1.** Divide the second row by -4: $$\begin{pmatrix} 1 & 1 & -1 & | & 11 \\ 0 & 1 & -1 & | & 13 \\ 0 & -5 & 3 & | & -85 \end{pmatrix}$$ **Step 4: Make the elements above and below the leading 1 in the second column zero.** Subtract the second row from the first row: $$\text{Row 1} = \text{Row 1} - \text{Row 2}$$ $$\begin{pmatrix} 1 & 0 & 0 & | & -2 \\ 0 & 1 & -1 & | & 13 \\ 0 & -5 & 3 & | & -85 \end{pmatrix}$$ Add 5 times the second row to the third row: $$\text{Row 3} = \text{Row 3} + 5 \cdot \text{Row 2}$$ $$\begin{pmatrix} 1 & 0 & 0 & | & -2 \\ 0 & 1 & -1 & | & 13 \\ 0 & 0 & -2 & | & -20 \end{pmatrix}$$ **Step 5: Make the element in the third row, third column a 1.** Divide the third row by -2: $$\begin{pmatrix} 1 & 0 & 0 & | & -2 \\ 0 & 1 & -1 & | & 13 \\ 0 & 0 & 1 & | & 10 \end{pmatrix}$$ **Step 6: Make the elements above the leading 1 in the third column zero.** Add the third row to the second row: $$\text{Row 2} = \text{Row 2} + \text{Row 3}$$ $$\begin{pmatrix} 1 & 0 & 0 & | & -2 \\ 0 & 1 & 0 & | & 23 \\ 0 & 0 & 1 & | & 10 \end{pmatrix}$$ The final augmented matrix represents the solution to the system: $$\begin{cases} x = -2 \\ y = 23 \\ z = 10 \end{cases}$$ So, the solution to the system is $$x = -2$$, $$y = 23$$, and $$z = 10$$. If you have any questions or need further details, feel free to ask! Here are 5 questions you might find interesting: 1. How do you solve a system of equations using the elimination method? 2. What are the advantages of the Gauss-Jordan method over other methods? 3. Can you explain how to interpret the reduced row echelon form of a matrix? 4. How do you determine the rank of a matrix? 5. What are some applications of solving systems of linear equations in real life? **Tip:** When solving systems of equations, always check your final answers by substituting them back into the original equations to ensure they satisfy all equations.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Gauss-Jordan Method

Formulas

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Theorems

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Suitable Grade Level

Advanced High School

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