Math Problem Statement
Use the Gauss-Jordan method to solve the system:
\(x - 2y + z = 5\)
\(x + y - 5z = -1\)
\(3x - 5y + 2z = 16\)
Solution
To solve this system of equations using the Gauss-Jordan method, we will transform the augmented matrix into reduced row echelon form. The system given is:
x - 2y + z = 5 \\ x + y - 5z = -1 \\ 3x - 5y + 2z = 16 \end{cases}$$ ### Step 1: Write the Augmented Matrix $$\begin{bmatrix} 1 & -2 & 1 & | & 5 \\ 1 & 1 & -5 & | & -1 \\ 3 & -5 & 2 & | & 16 \end{bmatrix}$$ ### Step 2: Make the First Element in the First Column a Leading 1 The first row already has a 1 in the first column, so we’ll use it as our pivot. ### Step 3: Eliminate the First Element in the Other Rows - Subtract Row 1 from Row 2 to make the first element of Row 2 zero: $$R_2 = R_2 - R_1 \Rightarrow [0, 3, -6, | -6]$$ - Subtract 3 times Row 1 from Row 3: $$R_3 = R_3 - 3 \cdot R_1 \Rightarrow [0, 1, -1, | 1]$$ The matrix is now: $$\begin{bmatrix} 1 & -2 & 1 & | & 5 \\ 0 & 3 & -6 & | & -6 \\ 0 & 1 & -1 & | & 1 \end{bmatrix}$$ ### Step 4: Make the Second Element in the Second Row a Leading 1 Divide Row 2 by 3: $$R_2 = \frac{1}{3} R_2 \Rightarrow [0, 1, -2, | -2]$$ The matrix is now: $$\begin{bmatrix} 1 & -2 & 1 & | & 5 \\ 0 & 1 & -2 & | -2 \\ 0 & 1 & -1 & | 1 \end{bmatrix}$$ ### Step 5: Eliminate the Second Element in the Other Rows - Add 2 times Row 2 to Row 1 to make the second element in Row 1 zero: $$R_1 = R_1 + 2 \cdot R_2 \Rightarrow [1, 0, -3, | 1]$$ - Subtract Row 2 from Row 3: $$R_3 = R_3 - R_2 \Rightarrow [0, 0, 1, | 3]$$ The matrix is now: $$\begin{bmatrix} 1 & 0 & -3 & | & 1 \\ 0 & 1 & -2 & | & -2 \\ 0 & 0 & 1 & | & 3 \end{bmatrix}$$ ### Step 6: Make the Third Element in the Third Row a Leading 1 Row 3 already has a leading 1 in the third column. ### Step 7: Eliminate the Third Element in the Other Rows - Add 3 times Row 3 to Row 1: $$R_1 = R_1 + 3 \cdot R_3 \Rightarrow [1, 0, 0, | 10]$$ - Add 2 times Row 3 to Row 2: $$R_2 = R_2 + 2 \cdot R_3 \Rightarrow [0, 1, 0, | 4]$$ The matrix is now in reduced row echelon form: $$\begin{bmatrix} 1 & 0 & 0 & | & 10 \\ 0 & 1 & 0 & | & 4 \\ 0 & 0 & 1 & | & 3 \end{bmatrix}$$ ### Solution The solution to the system is: $$x = 10, \quad y = 4, \quad z = 3$$ Would you like a more detailed explanation of any of these steps, or do you have any questions? --- ### Related Questions 1. How would you solve a system of equations using the Gaussian elimination method instead of Gauss-Jordan? 2. What are the differences between Gaussian elimination and Gauss-Jordan elimination? 3. How would you check if the solution obtained is correct for the given system of equations? 4. How can you interpret the results of a system of equations if you find infinitely many solutions? 5. How would the method change if one of the rows was entirely zeros? ### Tip When solving a system with the Gauss-Jordan method, always aim to achieve a diagonal of 1s with zeros above and below each leading 1. This helps you reach the solution more systematically.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Gauss-Jordan Elimination
Systems of Linear Equations
Formulas
Row operations to achieve reduced row echelon form (RREF)
Theorems
Gauss-Jordan Elimination Method
Suitable Grade Level
Grades 10-12
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