Math Problem Statement
2x-3y-5z=1, 3x+y-2z=-4, x-2y+z=5 by Gaussian method
Solution
To solve the system of linear equations using the Gaussian elimination method, we follow these steps:
Given system of equations:
Step 1: Write the system as an augmented matrix.
2 & -3 & -5 & | & 1 \\ 3 & 1 & -2 & | & -4 \\ 1 & -2 & 1 & | & 5 \\ \end{pmatrix}$$ **Step 2: Perform row operations to get a triangular (upper triangular) form.** We begin by making the coefficient of $$x$$ in the first column of the first row a leading 1. 1. $$R_1 \leftarrow R_1 / 2$$ $$\begin{pmatrix} 1 & -1.5 & -2.5 & | & 0.5 \\ 3 & 1 & -2 & | & -4 \\ 1 & -2 & 1 & | & 5 \\ \end{pmatrix}$$ **Step 3: Eliminate the first variable $$x$$ from the second and third rows.** 2. $$R_2 \leftarrow R_2 - 3R_1$$ and $$R_3 \leftarrow R_3 - R_1$$ $$\begin{pmatrix} 1 & -1.5 & -2.5 & | & 0.5 \\ 0 & 5.5 & 5.5 & | & -5.5 \\ 0 & -0.5 & 3.5 & | & 4.5 \\ \end{pmatrix}$$ **Step 4: Make the coefficient of $$y$$ in the second row a leading 1.** 3. $$R_2 \leftarrow R_2 / 5.5$$ $$\begin{pmatrix} 1 & -1.5 & -2.5 & | & 0.5 \\ 0 & 1 & 1 & | & -1 \\ 0 & -0.5 & 3.5 & | & 4.5 \\ \end{pmatrix}$$ **Step 5: Eliminate the second variable $$y$$ from the third row.** 4. $$R_3 \leftarrow R_3 + 0.5R_2$$ $$\begin{pmatrix} 1 & -1.5 & -2.5 & | & 0.5 \\ 0 & 1 & 1 & | & -1 \\ 0 & 0 & 4 & | & 4 \\ \end{pmatrix}$$ **Step 6: Solve the triangular system.** Now we can solve this system using back substitution. From $$R_3$$: $$4z = 4 \implies z = 1$$ From $$R_2$$: $$y + z = -1 \implies y + 1 = -1 \implies y = -2$$ From $$R_1$$: $$x - 1.5y - 2.5z = 0.5 \implies x - 1.5(-2) - 2.5(1) = 0.5 \implies x + 3 - 2.5 = 0.5 \implies x = 0$$ **Solution:** $$x = 0, \quad y = -2, \quad z = 1$$ This is the solution to the system of equations. Do you want any further details or have any questions? --- Here are 5 related questions to consider: 1. How would you solve the system using the Gauss-Jordan elimination method? 2. What changes in the Gaussian elimination process when the system has infinite solutions? 3. Can Gaussian elimination be used for systems with no solutions? 4. How would you solve the system if it were a 4x4 matrix instead? 5. What are the advantages of using Gaussian elimination over substitution? **Tip:** Gaussian elimination is more efficient for large systems of equations than substitution or elimination methods by hand.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Systems of Equations
Gaussian Elimination
Formulas
Row operations for matrix manipulation
Back substitution in triangular form
Theorems
Gaussian Elimination Theorem
Suitable Grade Level
Grades 10-12
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