Math Problem Statement

реши матричное уравнение методом гаусса

Solution

To solve the matrix equation $A \cdot X = B$ using the Gaussian elimination method, the steps involve:

Augmenting matrix $A$ with $B$ to create an augmented matrix $[A|B]$ .
Using row operations to reduce $[A|B]$ to row echelon form.
Back-substituting to find the solution $X$ .

Let me extract the matrices $A$ and $B$ from the image and then proceed with the solution. PIL import Image

Load and display the uploaded image to verify contents

image_path = "/mnt/data/1736859170966zanqe4xc.jpg" image = Image.open(image_path) image.show()I have opened the uploaded image to verify the content of matrices $A$ and $B$ . Let me proceed to extract and solve the matrix equation.The matrices from the problem are:

\begin{pmatrix} 5 & 6 & 2 & 2 \\ 11 & 16 & 6 & 6 \\ -6 & -12 & -3 & -4 \\ -20 & -28 & -4 & -7 \end{pmatrix}, \quad B = \begin{pmatrix} -4 & -6 \\ -6 & -16 \\ -2 & 10 \\ 6 & 22 \end{pmatrix}.$$ Now, we'll solve the equation $$A \cdot X = B$$ using the Gaussian elimination method. The solution will involve forming the augmented matrix $$[A|B]$$ and reducing it. Let's begin! numpy as np # Define matrices A and B A = np.array([ [5, 6, 2, 2], [11, 16, 6, 6], [-6, -12, -3, -4], [-20, -28, -4, -7] ], dtype=float) B = np.array([ [-4, -6], [-6, -16], [-2, 10], [6, 22] ], dtype=float) # Augment A with B augmented_matrix = np.hstack((A, B)) # Perform Gaussian elimination def gaussian_elimination(aug_matrix): rows, cols = aug_matrix.shape for i in range(rows): # Make the diagonal element 1 (if possible) if aug_matrix[i, i] != 0: aug_matrix[i] = aug_matrix[i] / aug_matrix[i, i] # Eliminate elements below the pivot for j in range(i + 1, rows): aug_matrix[j] -= aug_matrix[i] * aug_matrix[j, i] # Back substitution to eliminate elements above the pivot for i in range(rows - 1, -1, -1): for j in range(i - 1, -1, -1): aug_matrix[j] -= aug_matrix[i] * aug_matrix[j, i] return aug_matrix # Apply Gaussian elimination reduced_matrix = gaussian_elimination(augmented_matrix.copy()) # Extract the solution matrix X from the reduced matrix X = reduced_matrix[:, -B.shape[1]:] reduced_matrix, XAfter applying Gaussian elimination, the reduced matrix is: $$\begin{bmatrix} 1 & 0 & 0 & 0 & -2 & 0 \\ 0 & 1 & 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & 0 & -2 & -2 \\ 0 & 0 & 0 & 1 & 2 & 2 \end{bmatrix}.$$ The solution matrix $$X$$ is extracted as: $$X = \begin{bmatrix} -2 & 0 \\ 1 & -1 \\ -2 & -2 \\ 2 & 2 \end{bmatrix}.$$ This represents the solution to $$A \cdot X = B$$. ### Questions to Explore: 1. What is the significance of Gaussian elimination in solving matrix equations? 2. How would the process change if the matrix $$A$$ were singular (non-invertible)? 3. Can the solution $$X$$ be verified by substituting back into the original equation? 4. How does Gaussian elimination handle systems with no or infinite solutions? 5. What are alternative methods for solving matrix equations, and how do they compare? ### Tip: When solving matrix equations, always check the determinant of $$A$$. If it’s zero, $$A$$ is singular, and the system might not have a unique solution.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Equations
Gaussian Elimination

Formulas

Augmented matrix representation [A|B]
Row operations for Gaussian elimination
Back-substitution to solve linear systems

Theorems

Gaussian Elimination Method

Suitable Grade Level

Undergraduate Mathematics or Advanced High School (Grades 11-12)

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