Math Problem Statement
Solve the system of linear equations: x₁ - 3x₂ + 4x₃ = -4, 3x₁ - 7x₂ + 7x₃ = -8, -4x₁ + 6x₂ - x₃ = 7.
Solution
The system of linear equations provided is:
1. \quad &x_1 - 3x_2 + 4x_3 = -4 \\ 2. \quad &3x_1 - 7x_2 + 7x_3 = -8 \\ 3. \quad &-4x_1 + 6x_2 - x_3 = 7 \end{aligned}$$ We can solve this system using methods such as substitution, elimination, or matrix techniques like Gaussian elimination. I'll proceed with Gaussian elimination. ### Step 1: Write the augmented matrix $$\begin{pmatrix} 1 & -3 & 4 & | & -4 \\ 3 & -7 & 7 & | & -8 \\ -4 & 6 & -1 & | & 7 \end{pmatrix}$$ ### Step 2: Perform row operations to reach row-echelon form - Let's make the pivot of the first column 1, which is already the case in row 1. - Now, eliminate the values below this pivot in rows 2 and 3 by subtracting suitable multiples of row 1. 1. $$R_2 \leftarrow R_2 - 3R_1$$ $$R_2 = \begin{pmatrix} 3 & -7 & 7 & | & -8 \end{pmatrix} - 3 \times \begin{pmatrix} 1 & -3 & 4 & | & -4 \end{pmatrix} = \begin{pmatrix} 0 & 2 & -5 & | & 4 \end{pmatrix}$$ 2. $$R_3 \leftarrow R_3 + 4R_1$$ $$R_3 = \begin{pmatrix} -4 & 6 & -1 & | & 7 \end{pmatrix} + 4 \times \begin{pmatrix} 1 & -3 & 4 & | & -4 \end{pmatrix} = \begin{pmatrix} 0 & -6 & 15 & | & -9 \end{pmatrix}$$ The matrix now becomes: $$\begin{pmatrix} 1 & -3 & 4 & | & -4 \\ 0 & 2 & -5 & | & 4 \\ 0 & -6 & 15 & | & -9 \end{pmatrix}$$ ### Step 3: Eliminate below the second pivot We want to eliminate the value below the second pivot, which is the -6 in row 3. First, let's make the second pivot equal to 1 by dividing row 2 by 2. $$R_2 \leftarrow \frac{1}{2} R_2 = \begin{pmatrix} 0 & 1 & -\frac{5}{2} & | & 2 \end{pmatrix}$$ Now eliminate the -6 in row 3: $$R_3 \leftarrow R_3 + 6R_2 = \begin{pmatrix} 0 & -6 & 15 & | & -9 \end{pmatrix} + 6 \times \begin{pmatrix} 0 & 1 & -\frac{5}{2} & | & 2 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & | & 3 \end{pmatrix}$$ The matrix becomes: $$\begin{pmatrix} 1 & -3 & 4 & | & -4 \\ 0 & 1 & -\frac{5}{2} & | & 2 \\ 0 & 0 & 0 & | & 3 \end{pmatrix}$$ ### Step 4: Interpretation The third row becomes: $$0 = 3$$ This is a contradiction, meaning the system of equations has **no solution**. Would you like further details on this process, or do you have any questions? Here are 5 related questions you might consider: 1. What are the different methods to solve a system of linear equations? 2. How does Gaussian elimination work for systems with unique, infinite, or no solutions? 3. What is the difference between row-echelon and reduced row-echelon form? 4. Can matrix inverses be used to solve this system? 5. What are consistent and inconsistent systems in linear algebra? **Tip:** Always check for contradictions when solving systems of equations, as they indicate no solutions or inconsistencies.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
System of Linear Equations
Gaussian Elimination
Formulas
Gaussian elimination
Theorems
No solution theorem due to inconsistency
Suitable Grade Level
Grades 10-12
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