Math Problem Statement
solve by row echelon form
Solution
Let's solve the system of linear equations given in the image using row echelon form. The system is:
x + y + z &= 5 \quad \text{(Equation 1)} \\ 2x + 2y + z &= 6 \quad \text{(Equation 2)} \\ 4x + 3y + 6z &= 12 \quad \text{(Equation 3)} \end{aligned}$$ ### Step 1: Write the Augmented Matrix First, express the system as an augmented matrix: $$\begin{pmatrix} 1 & 1 & 1 & | & 5 \\ 2 & 2 & 1 & | & 6 \\ 4 & 3 & 6 & | & 12 \end{pmatrix}$$ ### Step 2: Perform Row Operations to Achieve Row Echelon Form **Eliminate the entries below the first leading 1:** - Subtract 2 times the first row from the second row ($$R_2 \rightarrow R_2 - 2R_1$$): $$R_2 \rightarrow \begin{pmatrix} 2 & 2 & 1 & | & 6 \end{pmatrix} - 2\begin{pmatrix} 1 & 1 & 1 & | & 5 \end{pmatrix} = \begin{pmatrix} 0 & 0 & -1 & | & -4 \end{pmatrix}$$ - Subtract 4 times the first row from the third row ($$R_3 \rightarrow R_3 - 4R_1$$): $$R_3 \rightarrow \begin{pmatrix} 4 & 3 & 6 & | & 12 \end{pmatrix} - 4\begin{pmatrix} 1 & 1 & 1 & | & 5 \end{pmatrix} = \begin{pmatrix} 0 & -1 & 2 & | & -8 \end{pmatrix}$$ The matrix now looks like: $$\begin{pmatrix} 1 & 1 & 1 & | & 5 \\ 0 & 0 & -1 & | & -4 \\ 0 & -1 & 2 & | & -8 \end{pmatrix}$$ **Eliminate the entry below the second leading entry:** - Add the third row to the second row ($$R_3 \rightarrow R_3 + R_2$$): $$R_3 \rightarrow \begin{pmatrix} 0 & -1 & 2 & | & -8 \end{pmatrix} + \begin{pmatrix} 0 & 0 & -1 & | & -4 \end{pmatrix} = \begin{pmatrix} 0 & -1 & 1 & | & -12 \end{pmatrix}$$ Now the matrix is in row echelon form: $$\begin{pmatrix} 1 & 1 & 1 & | & 5 \\ 0 & 0 & -1 & | & -4 \\ 0 & -1 & 1 & | & -12 \end{pmatrix}$$ ### Step 3: Solve the System by Back-Substitution Start from the bottom row and work upwards: 1. **Third row**: $$-y + z = -12$$ → $$z = -12 + y$$ 2. **Second row**: $$-z = -4$$ → $$z = 4$$ Substitute $$z = 4$$ into the third row equation to find $$y$$: $$-y + 4 = -12 \quad \Rightarrow \quad y = 16$$ 3. **First row**: $$x + y + z = 5$$ Substitute $$y = 16$$ and $$z = 4$$: $$x + 16 + 4 = 5 \quad \Rightarrow \quad x = -15$$ ### Final Solution: $$x = -15, \quad y = 16, \quad z = 4$$ ### Would you like any further details or explanations on any steps? --- Here are 8 related questions that expand on this topic: 1. How can you determine if a system of linear equations has no solution or infinitely many solutions using row echelon form? 2. What are the key differences between Gaussian elimination and Gauss-Jordan elimination? 3. How does the concept of linear independence relate to the solutions of systems of linear equations? 4. Can row echelon form be used to solve underdetermined systems (fewer equations than unknowns)? 5. What is the geometric interpretation of the solution set of a system of linear equations? 6. How can matrix rank be used to determine the number of solutions to a system of linear equations? 7. In what situations would row echelon form fail to provide a unique solution? 8. What are some applications of solving linear systems using row echelon form in engineering and science? --- **Tip:** Always check your final solution by substituting the values back into the original equations to ensure consistency.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Systems of Linear Equations
Row Echelon Form
Matrix Operations
Formulas
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Theorems
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Suitable Grade Level
Advanced High School or College Level