Math Problem Statement
Solve the equation using Guassian elimination method
X1+2x2+3x3=4
X1+3x2+5x3=1
X1+5x2-12x3=2
Solution
To solve the system of linear equations using Gaussian elimination, we first represent the system in augmented matrix form, then perform row operations to achieve a row-echelon form, and finally back-substitute to find the solution.
Step 1: Write the augmented matrix
The given system of equations is:
1. & \quad x_1 + 2x_2 + 3x_3 = 4 \\ 2. & \quad x_1 + 3x_2 + 5x_3 = 1 \\ 3. & \quad x_1 + 5x_2 - 12x_3 = 2 \\ \end{aligned}$$ This can be written as an augmented matrix: $$\begin{pmatrix} 1 & 2 & 3 & | & 4 \\ 1 & 3 & 5 & | & 1 \\ 1 & 5 & -12 & | & 2 \end{pmatrix}$$ ### Step 2: Perform row operations **First, eliminate $$x_1$$ from rows 2 and 3.** - Subtract Row 1 from Row 2: $$R2 = R2 - R1 \Rightarrow (1 - 1) & \quad (3 - 2) & \quad (5 - 3) & \quad (1 - 4) \\ \Rightarrow \begin{pmatrix} 0 & 1 & 2 & | & -3 \end{pmatrix}$$ - Subtract Row 1 from Row 3: $$R3 = R3 - R1 \Rightarrow (1 - 1) & \quad (5 - 2) & \quad (-12 - 3) & \quad (2 - 4) \\ \Rightarrow \begin{pmatrix} 0 & 3 & -15 & | & -2 \end{pmatrix}$$ So, the augmented matrix becomes: $$\begin{pmatrix} 1 & 2 & 3 & | & 4 \\ 0 & 1 & 2 & | & -3 \\ 0 & 3 & -15 & | & -2 \end{pmatrix}$$ **Next, eliminate the coefficient of $$x_2$$ in Row 3.** - Subtract 3 times Row 2 from Row 3: $$R3 = R3 - 3R2 \Rightarrow \begin{pmatrix} 0 & 3 & -15 & | & -2 \end{pmatrix} - 3\begin{pmatrix} 0 & 1 & 2 & | & -3 \end{pmatrix} \Rightarrow \begin{pmatrix} 0 & 0 & -21 & | & 7 \end{pmatrix}$$ The matrix now looks like this: $$\begin{pmatrix} 1 & 2 & 3 & | & 4 \\ 0 & 1 & 2 & | & -3 \\ 0 & 0 & -21 & | & 7 \end{pmatrix}$$ ### Step 3: Back-substitution Now that the matrix is in row-echelon form, we can solve for the variables starting from the bottom: 1. From Row 3: $$-21x_3 = 7$$ $$x_3 = \frac{7}{-21} = -\frac{1}{3}$$ 2. From Row 2: $$x_2 + 2x_3 = -3$$ $$x_2 + 2\left(-\frac{1}{3}\right) = -3 \Rightarrow x_2 - \frac{2}{3} = -3 \Rightarrow x_2 = -3 + \frac{2}{3} = -\frac{7}{3}$$ 3. From Row 1: $$x_1 + 2x_2 + 3x_3 = 4$$ $$x_1 + 2\left(-\frac{7}{3}\right) + 3\left(-\frac{1}{3}\right) = 4 \Rightarrow x_1 - \frac{14}{3} - 1 = 4 \Rightarrow x_1 = 4 + \frac{14}{3} + 1 = \frac{31}{3}$$ ### Final Solution The solution to the system of equations is: $$x_1 = \frac{31}{3}, \quad x_2 = -\frac{7}{3}, \quad x_3 = -\frac{1}{3}$$ Would you like to go into more details or do you have any questions? Here are some related problems: 1. Solve a 3x3 system with a different set of coefficients using Gaussian elimination. 2. How would you handle a system with no solutions or infinite solutions using Gaussian elimination? 3. Compare Gaussian elimination with the Gauss-Jordan method. 4. Solve a 4x4 system using Gaussian elimination. 5. What is the significance of the determinant in solving systems of equations? **Tip:** Always check the determinant of the coefficient matrix before applying Gaussian elimination to ensure the system has a unique solution.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
Formulas
Row Operations
Back-Substitution
Augmented Matrix Representation
Theorems
Gaussian Elimination
Suitable Grade Level
Grades 10-12
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