Math Problem Statement
Solve the following system of linear equations using Gaussian elimination:
2x1 + 3x2 - x3 = 5
4x1 + x2 + 2x3 = 11
-2x1 + 5x2 + 3x3 = 0
Solution
The system of linear equations given in the image is:
1) &\ 2x_1 + 3x_2 - x_3 = 5 \\ 2) &\ 4x_1 + x_2 + 2x_3 = 11 \\ 3) &\ -2x_1 + 5x_2 + 3x_3 = 0 \\ \end{aligned}$$ The problem asks to solve this system using Gaussian elimination, and three steps (Part 1, Part 2, Part 3) are listed: ### Part 1: $$R_2 \leftarrow R_2 - 2 \times R_1$$ This step performs an elementary row operation to eliminate the $$x_1$$ term in the second row. Let's calculate this: - Multiply row 1 by 2: $$2(2x_1 + 3x_2 - x_3 = 5)$$ gives $$4x_1 + 6x_2 - 2x_3 = 10$$. - Subtract this from row 2: $$(4x_1 + x_2 + 2x_3 = 11) - (4x_1 + 6x_2 - 2x_3 = 10)$$ This results in: $$0x_1 - 5x_2 + 4x_3 = 1$$ So the new system is: $$\begin{aligned} 2x_1 + 3x_2 - x_3 &= 5 \\ -5x_2 + 4x_3 &= 1 \\ -2x_1 + 5x_2 + 3x_3 &= 0 \\ \end{aligned}$$ ### Part 2: $$R_3 \leftarrow R_3 + R_1$$ This step eliminates $$x_1$$ from the third row by adding row 1 to row 3. - Add row 1 to row 3: $$(-2x_1 + 5x_2 + 3x_3 = 0) + (2x_1 + 3x_2 - x_3 = 5)$$ This gives: $$0x_1 + 8x_2 + 2x_3 = 5$$ So the system is now: $$\begin{aligned} 2x_1 + 3x_2 - x_3 &= 5 \\ -5x_2 + 4x_3 &= 1 \\ 8x_2 + 2x_3 &= 5 \\ \end{aligned}$$ ### Part 3: $$R_3 \leftarrow R_3 - \frac{8}{5} \times R_2$$ This step eliminates the $$x_2$$ term from the third row by scaling row 2 and subtracting it from row 3. - Multiply row 2 by $$\frac{8}{5}$$: $$\frac{8}{5}(-5x_2 + 4x_3 = 1) \Rightarrow -8x_2 + \frac{32}{5}x_3 = \frac{8}{5}$$ - Subtract this from row 3: $$(8x_2 + 2x_3 = 5) - (-8x_2 + \frac{32}{5}x_3 = \frac{8}{5})$$ This results in: $$0x_2 + \left(2 - \frac{32}{5}\right)x_3 = 5 - \frac{8}{5}$$ Simplifying: $$0x_2 - \frac{22}{5}x_3 = \frac{17}{5}$$ Multiply by $$-\frac{5}{22}$$ to solve for $$x_3$$: $$x_3 = \frac{17}{22}$$ ### Back Substitution: Now, substitute $$x_3 = \frac{17}{22}$$ into the second equation: $$-5x_2 + 4 \times \frac{17}{22} = 1$$ This simplifies to: $$-5x_2 + \frac{68}{22} = 1 \Rightarrow -5x_2 = 1 - \frac{68}{22} = \frac{22}{22} - \frac{68}{22} = -\frac{46}{22}$$ Thus: $$x_2 = \frac{46}{110} = \frac{23}{55}$$ Finally, substitute $$x_2 = \frac{23}{55}$$ and $$x_3 = \frac{17}{22}$$ into the first equation: $$2x_1 + 3 \times \frac{23}{55} - \frac{17}{22} = 5$$ Simplifying: $$2x_1 + \frac{69}{55} - \frac{17}{22} = 5$$ Getting a common denominator: \[ 2x_1 + \frac{276}{220} - \frac{170}{220} = 5 \Rightarrow 2x_1 + \frac{106}{220Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
Formulas
Elementary Row Operations
Row Reduction Algorithm
Back Substitution
Theorems
Gaussian Elimination Method
Suitable Grade Level
College level (Linear Algebra)
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