Math Problem Statement

Use Gaussian Elimination to solve the linear system by finding an equivalent system in triangular form. 3x1 + 4x2 + 2x3 = 10 x1 + x3 = 3 12x2 - 4x3 = 16 The unique solution to the system is (x1, x2, x3) = [ , , ].

Solution

To solve this system using Gaussian elimination, we aim to put it into an upper triangular form. Let's go through each step.

Step 1: Set up the augmented matrix

The system of equations is:

3x_1 + 4x_2 + 2x_3 = 10 \\ x_1 + x_3 = 3 \\ 12x_2 - 4x_3 = 16 \end{cases}$$ The augmented matrix is: $$\begin{pmatrix} 3 & 4 & 2 & | & 10 \\ 1 & 0 & 1 & | & 3 \\ 0 & 12 & -4 & | & 16 \end{pmatrix}$$ ### Step 2: Make the first entry of the first column into a leading 1 Divide the first row by 3: $$\begin{pmatrix} 1 & \frac{4}{3} & \frac{2}{3} & | & \frac{10}{3} \\ 1 & 0 & 1 & | & 3 \\ 0 & 12 & -4 & | & 16 \end{pmatrix}$$ ### Step 3: Eliminate $$x_1$$ from the second row Subtract the first row from the second row to make the second row’s first column entry zero: $$\begin{pmatrix} 1 & \frac{4}{3} & \frac{2}{3} & | & \frac{10}{3} \\ 0 & -\frac{4}{3} & \frac{1}{3} & | & \frac{-1}{3} \\ 0 & 12 & -4 & | & 16 \end{pmatrix}$$ ### Step 4: Scale the second row to make the leading coefficient 1 Multiply the second row by $$-\frac{3}{4}$$: $$\begin{pmatrix} 1 & \frac{4}{3} & \frac{2}{3} & | & \frac{10}{3} \\ 0 & 1 & -\frac{1}{4} & | & \frac{1}{4} \\ 0 & 12 & -4 & | & 16 \end{pmatrix}$$ ### Step 5: Eliminate $$x_2$$ from the third row Subtract 12 times the second row from the third row: $$\begin{pmatrix} 1 & \frac{4}{3} & \frac{2}{3} & | & \frac{10}{3} \\ 0 & 1 & -\frac{1}{4} & | & \frac{1}{4} \\ 0 & 0 & -1 & | & 13 \end{pmatrix}$$ ### Step 6: Scale the third row to make the leading coefficient 1 Multiply the third row by $$-1$$: $$\begin{pmatrix} 1 & \frac{4}{3} & \frac{2}{3} & | & \frac{10}{3} \\ 0 & 1 & -\frac{1}{4} & | & \frac{1}{4} \\ 0 & 0 & 1 & | & -13 \end{pmatrix}$$ ### Step 7: Back-substitute to find the solutions 1. From the third row: $$x_3 = -13$$ 2. Substitute $$x_3 = -13$$ into the second row: $$x_2 - \frac{1}{4}(-13) = \frac{1}{4} \Rightarrow x_2 = 3$$ 3. Substitute $$x_2 = 3$$ and $$x_3 = -13$$ into the first row: $$x_1 + \frac{4}{3}(3) + \frac{2}{3}(-13) = \frac{10}{3} \Rightarrow x_1 = 6$$ ### Solution The unique solution is: $$(x_1, x_2, x_3) = (6, 3, -13)$$ Would you like further details or clarification on any step? Here are some related questions: 1. What is Gaussian elimination and how does it work in solving systems? 2. How does back-substitution work in a triangular system? 3. Can Gaussian elimination be used for non-square systems? 4. How do different row operations affect the solution of the system? 5. What are some common mistakes to avoid in Gaussian elimination? **Tip:** Always verify your solution by substituting it back into the original equations.

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Math Problem Analysis

Mathematical Concepts

Algebra
Linear Systems
Gaussian Elimination

Formulas

Gaussian Elimination steps for transforming to triangular form

Theorems

Gaussian Elimination method for solving linear equations

Suitable Grade Level

Grades 10-12