answer:1. Let us define the complex numbers alpha, beta, and gamma as follows: [ alpha = x (cos A + i sin A), quad beta = y (cos B + i sin B), quad gamma = z (cos C + i sin C) ] Given these, observe that: [ alpha^r = x^r (cos rA + i sin rA), quad beta^r = y^r (cos rB + i sin rB), quad gamma^r = z^r (cos rC + i sin rC) ] We also know: [ cos (A + B + C) = -1 quad text{and} quad sin (A + B + C) = 0 quad text{since} quad A + B + C = kpi quad (text{where} , k , text{is an integer}) ] 2. Now, F_r can be rewritten by summing up the imaginary parts of alpha^r, beta^r, and gamma^r: [ F_r = Im (alpha^r + beta^r + gamma^r) ] 3. By Vieta's formulas, since alpha, beta, gamma can be considered as the roots of a cubic polynomial, the polynomial coefficients satisfy: [ a = alpha + beta + gamma, quad b = alphabeta + betagamma + gammaalpha, quad c = alphabetagamma ] Given that F_1 = 0, a is purely real. 4. Evaluating the polynomial coefficients: [ b = frac{1}{2} left[(alpha + beta + gamma)^2 - (alpha^2 + beta^2 + gamma^2)right] ] Since F_2 = 0, b is purely real as well. 5. The product of the roots alpha, beta, gamma is: [ c = alpha beta gamma = x y z left(cos (A + B + C) + i sin (A + B + C)right) ] Since cos (A + B + C) = -1 and sin (A + B + C) = 0: [ c = -x y z ] and c is purely real. 6. Considering S_r = alpha^r + beta^r + gamma^r: [ S_1 = alpha + beta + gamma quad text{is real because} quad F_1 = 0 ] [ S_2 = alpha^2 + beta^2 + gamma^2 quad text{is real because} quad F_2 = 0 ] We already have S_0 = 3 as real. 7. Using induction: Suppose S_{r+1} is real if S_r, S_{r-1}, S_{r-2} are real: [ S_{r+1} - a S_r + b S_{r-1} - c S_{r-2} = 0 ] Simplifying and considering all coefficients are real: [ S_{r+1} = a S_r - b S_{r-1} + c S_{r-2} ] Hence, S_{r+1} is real if S_r, S_{r-1}, S_{r-2} are real. 8. By induction, for all positive integers r, S_r is real and consequently F_r = 0. Conclusion: (blacksquare)

answer:Here's a simple implementation of matrix multiplication in Python using NumPy for matrix operations: ```python import numpy as np def multiply_matrices(mat1, mat2): Multiply two 2D matrices together using standard matrix multiplication rules. Args: mat1 (np.ndarray): The first 2D matrix mat2 (np.ndarray): The second 2D matrix Returns: np.ndarray: The product of the two input matrices Raises: ValueError: If the input matrices cannot be multiplied (i.e., the number of columns in the first matrix does not match the number of rows in the second matrix) # Check if the matrices can be multiplied if mat1.shape[1] != mat2.shape[0]: raise ValueError(Matrices cannot be multiplied) # Use NumPy's built-in matmul function for matrix multiplication return np.matmul(mat1, mat2) # Example usage mat1 = np.array([[1, 2], [3, 4]]) mat2 = np.array([[5, 6], [7, 8]]) result = multiply_matrices(mat1, mat2) print(result) ``` Output: ``` [[19 22], [43 50]] ``` Explanation: 1. We first check if the input matrices can be multiplied by comparing the number of columns in the first matrix (`mat1.shape[1]`) with the number of rows in the second matrix (`mat2.shape[0]`). If they don't match, we raise a `ValueError`. 2. We use NumPy's `matmul` function to perform the matrix multiplication. This function takes care of the underlying multiplication and summation operations according to standard matrix multiplication rules. 3. The resulting product matrix is returned by the function. Note that this implementation relies on NumPy for matrix operations. If you want to implement matrix multiplication from scratch without using any libraries, you can use nested loops to iterate over the elements of the matrices and perform the multiplication and summation operations manually. However, this approach is generally less efficient and less convenient than using a library like NumPy.

answer:Here's the implementation of matrix multiplication from scratch without using any libraries: ```python def multiply_matrices(mat1, mat2): Multiply two 2D matrices together using standard matrix multiplication rules. Args: mat1 (list of lists): The first 2D matrix mat2 (list of lists): The second 2D matrix Returns: list of lists: The product of the two input matrices Raises: ValueError: If the input matrices cannot be multiplied (i.e., the number of columns in the first matrix does not match the number of rows in the second matrix) # Check if the matrices can be multiplied if len(mat1[0]) != len(mat2): raise ValueError(Matrices cannot be multiplied) # Initialize the result matrix with zeros result = [[0 for _ in range(len(mat2[0]))] for _ in range(len(mat1))] # Perform matrix multiplication for i in range(len(mat1)): for j in range(len(mat2[0])): for k in range(len(mat2)): result[i][j] += mat1[i][k] * mat2[k][j] return result # Example usage mat1 = [[1, 2], [3, 4]] mat2 = [[5, 6], [7, 8]] result = multiply_matrices(mat1, mat2) for row in result: print(row) ``` Output: ``` [19, 22] [43, 50] ``` Explanation: 1. We first check if the input matrices can be multiplied by comparing the number of columns in the first matrix (`len(mat1[0])`) with the number of rows in the second matrix (`len(mat2)`). If they don't match, we raise a `ValueError`. 2. We initialize the result matrix with zeros. The number of rows in the result matrix is the same as the number of rows in the first matrix (`len(mat1)`), and the number of columns is the same as the number of columns in the second matrix (`len(mat2[0])`). 3. We perform matrix multiplication using three nested loops: * The outer two loops iterate over the rows and columns of the result matrix, respectively. * The inner loop iterates over the columns of the first matrix (or the rows of the second matrix) and performs the multiplication and summation operations. 4. The resulting product matrix is returned by the function. Note that this implementation has a time complexity of O(n^3), where n is the number of rows (or columns) in the matrices. This can be optimized to O(n^2.81) using the Strassen algorithm or O(n^2.376) using the Coppersmith-Winograd algorithm, but these optimizations are more complex and typically not needed for most use cases.

answer:Here's the implementation of Strassen's algorithm for matrix multiplication: ```python import numpy as np def strassen_matrix_multiply(A, B): Multiply two 2^n x 2^n matrices using Strassen's algorithm. Args: A (np.ndarray): The first 2^n x 2^n matrix B (np.ndarray): The second 2^n x 2^n matrix Returns: np.ndarray: The product of the two input matrices Raises: ValueError: If the input matrices are not square or their size is not a power of 2 n = A.shape[0] # Check if the matrices are square and their size is a power of 2 if A.shape != (n, n) or B.shape != (n, n) or (n & (n - 1) != 0): raise ValueError(Matrices must be square and their size must be a power of 2) # Base case: 1x1 matrices if n == 1: return A * B # Split the matrices into quadrants mid = n // 2 A11, A12, A21, A22 = A[:mid, :mid], A[:mid, mid:], A[mid:, :mid], A[mid:, mid:] B11, B12, B21, B22 = B[:mid, :mid], B[:mid, mid:], B[mid:, :mid], B[mid:, mid:] # Recursively compute the intermediate matrices M1 = strassen_matrix_multiply(A11 + A22, B11 + B22) M2 = strassen_matrix_multiply(A21 + A22, B11) M3 = strassen_matrix_multiply(A11, B12 - B22) M4 = strassen_matrix_multiply(A22, B21 - B11) M5 = strassen_matrix_multiply(A11 + A12, B22) M6 = strassen_matrix_multiply(A21 - A11, B11 + B12) M7 = strassen_matrix_multiply(A12 - A22, B21 + B22) # Combine the intermediate matrices to get the final result C11 = M1 + M4 - M5 + M7 C12 = M3 + M5 C21 = M2 + M4 C22 = M1 - M2 + M3 + M6 # Combine the quadrants to get the final matrix C = np.vstack((np.hstack((C11, C12)), np.hstack((C21, C22)))) return C # Example usage A = np.array([[1, 2], [3, 4]]) B = np.array([[5, 6], [7, 8]]) result = strassen_matrix_multiply(A, B) print(result) ``` Output: ``` [[19 22], [43 50]] ``` Explanation: 1. We first check if the input matrices are square and their size is a power of 2. If not, we raise a `ValueError`. 2. We define the base case for 1x1 matrices, where the product is simply the element-wise multiplication. 3. We split the input matrices into quadrants and recursively compute seven intermediate matrices `M1` to `M7`. 4. We combine these intermediate matrices to get the final result matrix `C`, which is split into quadrants `C11`, `C12`, `C21`, and `C22`. 5. We combine the quadrants to get the final matrix `C`. Note that Strassen's algorithm has a time complexity of O(n^2.81), which is better than the standard matrix multiplication algorithm for large matrices. However, it requires more memory accesses and is generally slower for small matrices due to the overhead of recursive function calls.