Math Problem Statement

Solve the following system of linear equations by Gaussian elimination: 4x1+6x2+x3=2 (R1), 2x1+x2-4x3=3 (R2), 3x1-2x2+5x3=8 (R3)

Solution

To solve the system of linear equations using Gaussian elimination, we need to perform a series of row operations to convert the system into an upper triangular form, and then back-substitute to find the solutions.

Step 1: Write the system as an augmented matrix

Given the system:

4x_1 + 6x_2 + x_3 &= 2 &\quad \text{(R1)} \\ 2x_1 + x_2 - 4x_3 &= 3 &\quad \text{(R2)} \\ 3x_1 - 2x_2 + 5x_3 &= 8 &\quad \text{(R3)} \end{aligned}$$ The augmented matrix is: $$\begin{pmatrix} 4 & 6 & 1 & \vert & 2 \\ 2 & 1 & -4 & \vert & 3 \\ 3 & -2 & 5 & \vert & 8 \end{pmatrix}$$ ### Step 2: Perform row operations to create zeros below the pivot in the first column We will create zeros below the first pivot (which is 4 in R1): - **Operation 1:** Replace $$R2$$ with $$R2 - \frac{1}{2}R1$$ - **Operation 2:** Replace $$R3$$ with $$R3 - \frac{3}{4}R1$$ Performing these operations: $$R2 \rightarrow R2 - \frac{1}{2}R1 = \left( 2 - \frac{1}{2}(4),\ 1 - \frac{1}{2}(6),\ -4 - \frac{1}{2}(1),\ 3 - \frac{1}{2}(2) \right) = (0, -2, -\frac{9}{2}, 2)$$ $$R3 \rightarrow R3 - \frac{3}{4}R1 = \left( 3 - \frac{3}{4}(4),\ -2 - \frac{3}{4}(6),\ 5 - \frac{3}{4}(1),\ 8 - \frac{3}{4}(2) \right) = (0, -6.5, 4.25, 6.5)$$ The augmented matrix becomes: $$\begin{pmatrix} 4 & 6 & 1 & \vert & 2 \\ 0 & -2 & -\frac{9}{2} & \vert & 2 \\ 0 & -6.5 & 4.25 & \vert & 6.5 \end{pmatrix}$$ ### Step 3: Perform row operations to create zeros below the pivot in the second column We will create a zero below the pivot in the second column (which is now $$-2$$ in R2): - **Operation 3:** Replace $$R3$$ with $$R3 - \frac{-6.5}{-2}R2$$ $$R3 \rightarrow R3 - \frac{-6.5}{-2}R2 = \left( 0,\ -6.5 - 3.25(-2),\ 4.25 - 3.25\left(-\frac{9}{2}\right),\ 6.5 - 3.25(2) \right) = (0, 0, -12.375, 0)$$ The augmented matrix becomes: $$\begin{pmatrix} 4 & 6 & 1 & \vert & 2 \\ 0 & -2 & -\frac{9}{2} & \vert & 2 \\ 0 & 0 & -12.375 & \vert & 0 \end{pmatrix}$$ ### Step 4: Back substitution to find the solutions Now that we have an upper triangular matrix, we can back-substitute to find the values of $$x_1$$, $$x_2$$, and $$x_3$$. From the third row: $$-12.375x_3 = 0 \quad \Rightarrow \quad x_3 = 0$$ Substituting $$x_3 = 0$$ into the second row: $$-2x_2 - \frac{9}{2}(0) = 2 \quad \Rightarrow \quad -2x_2 = 2 \quad \Rightarrow \quad x_2 = -1$$ Substituting $$x_2 = -1$$ and $$x_3 = 0$$ into the first row: $$4x_1 + 6(-1) + 1(0) = 2 \quad \Rightarrow \quad 4x_1 - 6 = 2 \quad \Rightarrow \quad 4x_1 = 8 \quad \Rightarrow \quad x_1 = 2$$ ### Final Answer: The solution to the system is: $$x_1 = 2,\quad x_2 = -1,\quad x_3 = 0$$ Would you like more details or have any questions? --- Here are some related questions you might find interesting: 1. How can Gaussian elimination be used to determine if a system has no solutions? 2. What is the difference between Gaussian elimination and Gauss-Jordan elimination? 3. Can Gaussian elimination be applied to non-square matrices? 4. How does the determinant of the coefficient matrix affect the solutions of a system? 5. What are the computational complexities of Gaussian elimination? **Tip:** Always double-check each row operation when performing Gaussian elimination to avoid calculation errors.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Systems of Linear Equations
Gaussian Elimination

Formulas

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Theorems

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Suitable Grade Level

Undergraduate

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