Math Problem Statement
Find the solution to the system of equations using Gaussian elimination:
\( \begin{cases} 3x + y + 2z = -7 \\ x - 2y + z = 2 \\ x + y + 3z = -6 \end{cases} \)
Solution
To solve this system of linear equations using Gaussian elimination, we can transform the system into an augmented matrix and then perform row operations to bring it to row-echelon form. Here's the system of equations:
3x + y + 2z = -7 \\ x - 2y + z = 2 \\ x + y + 3z = -6 \end{cases}$$ ### Step 1: Set up the augmented matrix The augmented matrix for this system is: $$\begin{bmatrix} 3 & 1 & 2 & | & -7 \\ 1 & -2 & 1 & | & 2 \\ 1 & 1 & 3 & | & -6 \end{bmatrix}$$ ### Step 2: Make the first entry in the first column (pivot) equal to 1 To do this, we can swap row 1 and row 2: $$\begin{bmatrix} 1 & -2 & 1 & | & 2 \\ 3 & 1 & 2 & | & -7 \\ 1 & 1 & 3 & | & -6 \end{bmatrix}$$ ### Step 3: Eliminate entries below the pivot in the first column - Subtract $$3 \times \text{(Row 1)}$$ from Row 2: $$R_2 = R_2 - 3 \times R_1$$ This gives: $$\begin{bmatrix} 1 & -2 & 1 & | & 2 \\ 0 & 7 & -1 & | & -13 \\ 1 & 1 & 3 & | & -6 \end{bmatrix}$$ - Subtract $$1 \times \text{(Row 1)}$$ from Row 3: $$R_3 = R_3 - R_1$$ This gives: $$\begin{bmatrix} 1 & -2 & 1 & | & 2 \\ 0 & 7 & -1 & | & -13 \\ 0 & 3 & 2 & | & -8 \end{bmatrix}$$ ### Step 4: Make the second entry in the second column (pivot) equal to 1 Divide Row 2 by 7: $$R_2 = \frac{1}{7} R_2$$ This gives: $$\begin{bmatrix} 1 & -2 & 1 & | & 2 \\ 0 & 1 & -\frac{1}{7} & | & -\frac{13}{7} \\ 0 & 3 & 2 & | & -8 \end{bmatrix}$$ ### Step 5: Eliminate entries below and above the pivot in the second column - Subtract $$3 \times \text{(Row 2)}$$ from Row 3: $$R_3 = R_3 - 3 \times R_2$$ This gives: $$\begin{bmatrix} 1 & -2 & 1 & | & 2 \\ 0 & 1 & -\frac{1}{7} & | & -\frac{13}{7} \\ 0 & 0 & \frac{23}{7} & | & -\frac{25}{7} \end{bmatrix}$$ - Add $$2 \times \text{(Row 2)}$$ to Row 1: $$R_1 = R_1 + 2 \times R_2$$ This gives: $$\begin{bmatrix} 1 & 0 & \frac{5}{7} & | & -\frac{12}{7} \\ 0 & 1 & -\frac{1}{7} & | & -\frac{13}{7} \\ 0 & 0 & \frac{23}{7} & | & -\frac{25}{7} \end{bmatrix}$$ ### Step 6: Make the third entry in the third column (pivot) equal to 1 Divide Row 3 by $$\frac{23}{7}$$: $$R_3 = \frac{7}{23} R_3$$ This gives: \[ \begin{bmatrix} 1 & 0 & \frac{5}{7} & | & -\frac{12}{7} \\ 0 & 1 & -\frac{1}{7} & | & -\frac{13}{7} \\ 0 & 0 & 1 & | & -\frac{25}{23} \times \frac{7}{23}Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Gaussian Elimination
Systems of Equations
Formulas
Gaussian elimination process
Row operations for matrix manipulation
Theorems
Gaussian elimination method for solving systems of equations
Suitable Grade Level
Grades 10-12