Math Problem Statement
Determine the values of x, y, and z for the following equations using Gauss elimination and Gauss-Jordan elimination: 2x + 3y + z = 7 3x + 4y − 2z = 4 5x − y + 3z = 10
Solution
We have the following system of equations:
1. \quad & 2x + 3y + z = 7 \\ 2. \quad & 3x + 4y - 2z = 4 \\ 3. \quad & 5x - y + 3z = 10 \end{aligned}$$ ### Step 1: Write the augmented matrix $$\begin{pmatrix} 2 & 3 & 1 & | & 7 \\ 3 & 4 & -2 & | & 4 \\ 5 & -1 & 3 & | & 10 \end{pmatrix}$$ ### Step 2: Gauss Elimination (Row Reduction to Row Echelon Form) We will eliminate the elements below the first pivot (2 in the first row, first column). #### Eliminate below $$a_{11}$$: 1. To make $$a_{21} = 0$$, replace $$R_2$$ with $$R_2 - \frac{3}{2}R_1$$: $$R_2 \rightarrow R_2 - \frac{3}{2}R_1 = \begin{pmatrix} 0 & \frac{1}{2} & -\frac{7}{2} & | & -\frac{7}{2} \end{pmatrix}$$ 2. To make $$a_{31} = 0$$, replace $$R_3$$ with $$R_3 - \frac{5}{2}R_1$$: $$R_3 \rightarrow R_3 - \frac{5}{2}R_1 = \begin{pmatrix} 0 & -\frac{17}{2} & \frac{1}{2} & | & \frac{5}{2} \end{pmatrix}$$ The augmented matrix is now: $$\begin{pmatrix} 2 & 3 & 1 & | & 7 \\ 0 & \frac{1}{2} & -\frac{7}{2} & | & -\frac{7}{2} \\ 0 & -\frac{17}{2} & \frac{1}{2} & | & \frac{5}{2} \end{pmatrix}$$ #### Eliminate below $$a_{22}$$: 1. First, multiply $$R_2$$ by 2 to simplify: $$R_2 \rightarrow 2R_2 = \begin{pmatrix} 0 & 1 & -7 & | & -7 \end{pmatrix}$$ 2. Now, eliminate the element below $$a_{22}$$. Replace $$R_3$$ with $$R_3 + 17R_2$$: $$R_3 \rightarrow R_3 + 17R_2 = \begin{pmatrix} 0 & 0 & -118 & | & -114 \end{pmatrix}$$ The augmented matrix is now: $$\begin{pmatrix} 2 & 3 & 1 & | & 7 \\ 0 & 1 & -7 & | & -7 \\ 0 & 0 & -118 & | & -114 \end{pmatrix}$$ ### Step 3: Back Substitution to Solve for $$z$$, $$y$$, and $$x$$ From the last row, we can solve for $$z$$: $$-118z = -114 \quad \Rightarrow \quad z = \frac{-114}{-118} = \frac{57}{59}$$ Substitute $$z = \frac{57}{59}$$ into the second equation to solve for $$y$$: $$y - 7\left(\frac{57}{59}\right) = -7 \quad \Rightarrow \quad y = -7 + 7\left(\frac{57}{59}\right) = \frac{-4}{59}$$ Substitute $$y = \frac{-4}{59}$$ and $$z = \frac{57}{59}$$ into the first equation to solve for $$x$$: $$2x + 3\left(\frac{-4}{59}\right) + \left(\frac{57}{59}\right) = 7$$ $$2x + \frac{-12 + 57}{59} = 7 \quad \Rightarrow \quad 2x + \frac{45}{59} = 7$$ $$2x = 7 - \frac{45}{59} = \frac{413}{59} \quad \Rightarrow \quad x = \frac{413}{118}$$ ### Final Solutions: $$x = \frac{413}{118}, \quad y = \frac{-4}{59}, \quad z = \frac{57}{59}$$ ### Gauss-Jordan Elimination Now we would proceed to the Gauss-Jordan method, reducing the matrix further to a reduced row echelon form, but the solution will still be: $$x = \frac{413}{118}, \quad y = \frac{-4}{59}, \quad z = \frac{57}{59}$$ --- Would you like more detailed steps, or any further explanation? ### Related Questions: 1. How does Gauss elimination compare with other methods like matrix inversion for solving systems? 2. What is the geometric interpretation of solving a system of linear equations like this one? 3. How does rounding affect the accuracy of solutions in systems of linear equations? 4. How would the process change if one of the equations were dependent on another? 5. What is the purpose of scaling rows in Gauss-Jordan elimination? ### Tip: When solving linear systems with larger matrices, it's often more efficient to use Gaussian elimination with partial pivoting to reduce rounding errors.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Systems of Linear Equations
Gauss Elimination
Gauss-Jordan Elimination
Formulas
Row Echelon Form Transformation
Back Substitution
Reduced Row Echelon Form Transformation
Theorems
Gaussian Elimination
Gauss-Jordan Elimination
Suitable Grade Level
Grades 10-12
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