Math Problem Statement
Answer the question
Use Gauss' Method to find the general solution of the system and give one particular solution for the system of linear equations.
Solution
We are tasked with solving the given system of linear equations using Gauss' Method (also called Gaussian elimination) and finding both the general solution and a particular solution.
The system of equations is as follows:
3x_1 + 4x_2 + x_3 + 2x_4 + x_5 &= 4 \\ 6x_1 + 8x_2 + 2x_3 + 5x_4 - x_5 &= 6 \\ 9x_1 + 12x_2 + 3x_3 + 10x_4 - 4x_5 &= 9 \end{aligned}$$ ### Step 1: Write the Augmented Matrix First, we write the augmented matrix corresponding to the system: $$\begin{pmatrix} 3 & 4 & 1 & 2 & 1 & | & 4 \\ 6 & 8 & 2 & 5 & -1 & | & 6 \\ 9 & 12 & 3 & 10 & -4 & | & 9 \end{pmatrix}$$ ### Step 2: Perform Gaussian Elimination Our goal is to reduce this matrix to row-echelon form using elementary row operations. - **R2** ← R2 - 2 * R1: $$\begin{pmatrix} 3 & 4 & 1 & 2 & 1 & | & 4 \\ 0 & 0 & 0 & 1 & -3 & | & -2 \\ 9 & 12 & 3 & 10 & -4 & | & 9 \end{pmatrix}$$ - **R3** ← R3 - 3 * R1: $$\begin{pmatrix} 3 & 4 & 1 & 2 & 1 & | & 4 \\ 0 & 0 & 0 & 1 & -3 & | & -2 \\ 0 & 0 & 0 & 4 & -7 & | & -3 \end{pmatrix}$$ - **R3** ← R3 - 4 * R2: $$\begin{pmatrix} 3 & 4 & 1 & 2 & 1 & | & 4 \\ 0 & 0 & 0 & 1 & -3 & | & -2 \\ 0 & 0 & 0 & 0 & 5 & | & 5 \end{pmatrix}$$ ### Step 3: Back-Substitution The matrix is now in row-echelon form. We can now solve the system using back-substitution. From the third row: $$5x_5 = 5 \implies x_5 = 1$$ From the second row: $$x_4 - 3x_5 = -2 \implies x_4 - 3(1) = -2 \implies x_4 = 1$$ From the first row: $$3x_1 + 4x_2 + x_3 + 2x_4 + x_5 = 4 \implies 3x_1 + 4x_2 + x_3 + 2(1) + 1 = 4$$ Simplifying: $$3x_1 + 4x_2 + x_3 = 1$$ ### Step 4: General Solution The variables $$x_1$$, $$x_2$$, and $$x_3$$ are free variables. Therefore, we can express the general solution as: $$x_1 = \frac{1 - 4x_2 - x_3}{3}, \quad x_2 = x_2, \quad x_3 = x_3, \quad x_4 = 1, \quad x_5 = 1$$ ### Step 5: Particular Solution To find a particular solution, we can set the free variables to zero (e.g., $$x_2 = 0$$ and $$x_3 = 0$$): $$x_1 = \frac{1}{3}, \quad x_2 = 0, \quad x_3 = 0, \quad x_4 = 1, \quad x_5 = 1$$ So, one particular solution is: $$\left( x_1, x_2, x_3, x_4, x_5 \right) = \left( \frac{1}{3}, 0, 0, 1, 1 \right)$$ --- Would you like further details on any part of this process? Here are five related questions to deepen your understanding: 1. How does Gaussian elimination differ from Gauss-Jordan elimination? 2. What are the properties of the augmented matrix in a consistent system? 3. Can a system with more variables than equations have a unique solution? 4. How can we verify if a system has infinitely many solutions? 5. What are the geometric interpretations of solutions to linear systems? **Tip:** When performing Gaussian elimination, aim to make one leading coefficient in each row equal to 1 for ease of computation.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Gaussian Elimination
Systems of Linear Equations
Formulas
Gaussian elimination steps to reduce the augmented matrix to row-echelon form
Theorems
Gaussian Elimination
Suitable Grade Level
Undergraduate Level