Math Problem Statement
Solve the system of equations using the Gauss elimination method: x + 3y - 2z = 1, 2x - 3y + z = 0, 5x + 2y - 5z = 1, 3x + 13y - 12z = 5
Solution
Let's solve the given system of equations using the Gauss elimination method:
The system of equations is:
Step 1: Write the augmented matrix
We can express the system as an augmented matrix:
1 & 3 & -2 & | & 1 \\ 2 & -3 & 1 & | & 0 \\ 5 & 2 & -5 & | & 1 \\ 3 & 13 & -12 & | & 5 \end{pmatrix}$$ ### Step 2: Perform row operations to get the matrix in row echelon form - First, we want to make the elements below the leading 1 in the first column zero. - To make the element in the second row, first column zero, perform the row operation $$R_2 \to R_2 - 2R_1$$. - To make the element in the third row, first column zero, perform the row operation $$R_3 \to R_3 - 5R_1$$. - To make the element in the fourth row, first column zero, perform the row operation $$R_4 \to R_4 - 3R_1$$. I'll perform these steps next and solve for the variables. numpy as np # Initial augmented matrix A = np.array([[1, 3, -2, 1], [2, -3, 1, 0], [5, 2, -5, 1], [3, 13, -12, 5]], dtype=float) # Perform row operations to eliminate the first column A[1] = A[1] - 2 * A[0] A[2] = A[2] - 5 * A[0] A[3] = A[3] - 3 * A[0] AAfter performing the first set of row operations, we get the matrix: $$\begin{pmatrix} 1 & 3 & -2 & | & 1 \\ 0 & -9 & 5 & | & -2 \\ 0 & -13 & 5 & | & -4 \\ 0 & 4 & -6 & | & 2 \end{pmatrix}$$ ### Step 3: Eliminate the second column elements below the leading $$-9$$ (in the second row) - Perform the row operation $$R_3 \to R_3 - \frac{13}{9} R_2$$ to eliminate the second column entry in row 3. - Perform the row operation $$R_4 \to R_4 - \frac{4}{9} R_2$$ to eliminate the second column entry in row 4. Let's apply these row operations and continue solving. Perform row operations to eliminate the second column A[2] = A[2] - (13/9) * A[1] A[3] = A[3] - (4/9) * A[1] AAfter applying the row operations to eliminate the second column elements, we get the matrix: $$\begin{pmatrix} 1 & 3 & -2 & | & 1 \\ 0 & -9 & 5 & | & -2 \\ 0 & 0 & -2.22 & | & -1.11 \\ 0 & 8 & -8.22 & | & 2.89 \end{pmatrix}$$ ### Step 4: Eliminate the third column element in the fourth row - Perform the row operation $$R_4 \to R_4 - \frac{8}{-2.22} R_3$$ to eliminate the third column entry in row 4. Let's apply this row operation and solve the system further. Perform row operation to eliminate the third column element in the fourth row A[3] = A[3] - (8 / -2.22) * A[2] AAfter applying the row operation to eliminate the third column element in the fourth row, we have the following matrix: $$\begin{pmatrix} 1 & 3 & -2 & | & 1 \\ 0 & -9 & 5 & | & -2 \\ 0 & 0 & -2.22 & | & -1.11 \\ 0 & 8 & -16.23 & | & -1.12 \end{pmatrix}$$ ### Step 5: Back-substitute to find the values of $$x$$, $$y$$, and $$z$$ Now, we can proceed with back-substitution to find the solutions for $$x$$, $$y$$, and $$z$$. Let's perform the necessary steps. Solving the system by back-substitution # We can extract the system of equations and use numpy's linear algebra solver # Augmented matrix without the last column A_reduced = A[:, :-1] # Right-hand side of the augmented matrix B = A[:, -1] # Solving the system solutions = np.linalg.solve(A_reduced, B) solutionsIt seems like I can’t do more advanced data analysis right now. Please try again later. Let me know if you would like me to continue with any other part of the solution or if you have a different question!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Gaussian Elimination
Formulas
Row operations to form upper triangular matrix
Back substitution to solve system
Theorems
Gauss Elimination Method
Suitable Grade Level
Undergraduate Mathematics or Advanced High School
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