Math Problem Statement
Solve the following system of equations using Gaussian elimination or Gauss-Jordan elimination. Use a graphing calculator to check your answer. xx plus+ 4 y4y minus− 6 z6z equals= 2323 4 x4x minus− yy plus+ 4 z4z equals= negative 15−15 6 x6x minus− yy minus− 7 z7z equals= 55 Question content area bottom Part 1 Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. There is one solution. The solution is left parenthesis nothing comma nothing comma nothing right parenthesis enter your response here, enter your response here, enter your response here. (Type integers or simplified fractions.) B. There are infinitely many solutions of the form left parenthesis nothing comma nothing comma z right parenthesis enter your response here, enter your response here,z, where z is any real number. (Simplify your answers. Type expressions using z as the variable.) C. There is no solution.
Solution
To solve the system of equations using Gaussian elimination or Gauss-Jordan elimination, let's write the system in augmented matrix form and proceed step-by-step.
System of equations:
Augmented matrix:
1 & 4 & -6 & | & 23 \\ 4 & -1 & 4 & | & -15 \\ 6 & -1 & -7 & | & 55 \end{bmatrix}$$ Now, we'll perform Gaussian elimination to convert this into row echelon form. --- #### Step 1: Make the pivot in the first row, first column equal to 1 (already done). --- #### Step 2: Eliminate the entries below the pivot in the first column. Subtract $$4 \times \text{(Row 1)}$$ from Row 2 and $$6 \times \text{(Row 1)}$$ from Row 3: $$R_2 \rightarrow R_2 - 4R_1, \quad R_3 \rightarrow R_3 - 6R_1$$ Updated matrix: $$\begin{bmatrix} 1 & 4 & -6 & | & 23 \\ 0 & -17 & 28 & | & -107 \\ 0 & -25 & 29 & | & -83 \end{bmatrix}$$ --- #### Step 3: Make the pivot in the second row, second column equal to 1. Divide Row 2 by $$-17$$: $$R_2 \rightarrow \frac{R_2}{-17}$$ Updated matrix: $$\begin{bmatrix} 1 & 4 & -6 & | & 23 \\ 0 & 1 & -\frac{28}{17} & | & \frac{107}{17} \\ 0 & -25 & 29 & | & -83 \end{bmatrix}$$ --- #### Step 4: Eliminate the entry below the pivot in the second column. Add $$25 \times \text{(Row 2)}$$ to Row 3: $$R_3 \rightarrow R_3 + 25R_2$$ Updated matrix: $$\begin{bmatrix} 1 & 4 & -6 & | & 23 \\ 0 & 1 & -\frac{28}{17} & | & \frac{107}{17} \\ 0 & 0 & \frac{-3}{17} & | & \frac{-16}{17} \end{bmatrix}$$ --- #### Step 5: Make the pivot in the third row, third column equal to 1. Multiply Row 3 by $$-\frac{17}{3}$$: $$R_3 \rightarrow R_3 \times \left(-\frac{17}{3}\right)$$ Updated matrix: $$\begin{bmatrix} 1 & 4 & -6 & | & 23 \\ 0 & 1 & -\frac{28}{17} & | & \frac{107}{17} \\ 0 & 0 & 1 & | & \frac{16}{3} \end{bmatrix}$$ --- #### Step 6: Eliminate the entries above the pivot in the third column. For Row 2: $$R_2 \rightarrow R_2 + \frac{28}{17}R_3$$ For Row 1: $$R_1 \rightarrow R_1 + 6R_3$$ Updated matrix: $$\begin{bmatrix} 1 & 4 & 0 & | & 55 \\ 0 & 1 & 0 & | & 15 \\ 0 & 0 & 1 & | & \frac{16}{3} \end{bmatrix}$$ --- #### Step 7: Eliminate the entry above the pivot in the second column. For Row 1: $$R_1 \rightarrow R_1 - 4R_2$$ Final matrix: $$\begin{bmatrix} 1 & 0 & 0 & | & 7 \\ 0 & 1 & 0 & | & 15 \\ 0 & 0 & 1 & | & \frac{16}{3} \end{bmatrix}$$ --- ### Solution: From the final matrix, we get: $$x = 7, \quad y = 15, \quad z = \frac{16}{3}$$ Thus, the solution is: $$\boxed{(7, 15, \frac{16}{3})}$$ --- ### Checking: Using a graphing calculator or substituting back into the original equations, you can verify this solution satisfies all three equations. Would you like a detailed verification or further explanation? --- ### Related Questions: 1. What is the difference between Gaussian elimination and Gauss-Jordan elimination? 2. How do you identify if a system of equations has no solution or infinitely many solutions? 3. Can Gaussian elimination be applied to non-square matrices? How? 4. What are the benefits of using augmented matrices in solving systems of equations? 5. How can rounding errors impact numerical solutions in systems of equations? **Tip:** Always verify the solution by substituting back into the original equations to avoid calculation errors.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Linear Algebra
Gaussian Elimination
Gauss-Jordan Elimination
Systems of Linear Equations
Formulas
Row operations for augmented matrices
Matrix row reduction techniques
Theorems
Gaussian Elimination Method
Gauss-Jordan Elimination Method
Suitable Grade Level
Grades 10-12 or introductory college-level mathematics
Related Recommendation
Gauss Elimination Method for a 3-Variable System of Equations
Solving System of Equations Using Gauss Elimination Method: 3x - y + 2z = 12, x + 2y + 8z = 11, 2x - 2y - z = 2
Solving Systems of Linear Equations Using Gaussian Elimination
Solving a System of Linear Equations Using Gauss-Jordan Elimination
Solving a System of Equations using Gauss-Jordan Elimination