Problem Analysis #9

[PYTHON] import numpy as np import sympy

def calculate_determinant(): # Construct the matrix n = 10 # Diagonal 20, off-diagonal 10 # A = 10 * I + 10 * J # Or simply construct it element by element A_list = [] for i in range(n): row = [] for j in range(n): if i == j: row.append(20) else: row.append(10) A_list.append(row)

# Use sympy for exact calculation
A_sympy = sympy.Matrix(A_list)
det_sympy = A_sympy.det()

# Use numpy for verification (though float precision might be an issue, for this size it should be fine)
A_np = np.array(A_list)
det_np = np.linalg.det(A_np)

print(f"Sympy determinant: {det_sympy}")
print(f"Numpy determinant: {det_np}")

# Theoretical calculation
# Eigenvalues are 10 + 10*10 = 110 (mult 1) and 10 (mult 9)
theoretical = 110 * (10**9)
print(f"Theoretical determinant: {theoretical}")

calculate_determinant() [/PYTHON] [THOUGHT] The matrix $A$ is a $10 \times 10$ matrix with diagonal entries 20 and off-diagonal entries 10. This can be written as $A = 10I + 10J$, where $I$ is the identity matrix and $J$ is the all-ones matrix. The eigenvalues of $J$ are $n=10$ (with multiplicity 1) and $0$ (with multiplicity $n-1=9$). Therefore, the eigenvalues of $A$ are: $\lambda_1 = 10 + 10(10) = 110$ (with multiplicity 1) $\lambda_2 = 10 + 10(0) = 10$ (with multiplicity 9) The determinant is the product of the eigenvalues: $\det(A) = 110 \times 10^9 = 11 \times 10^{10} = 110,000,000,000$. The Python code using sympy confirmed this exact integer value.

[/THOUGHT]

\boxed{110000000000}

The matrix $A$ is a $10 \times 10$ matrix with diagonal entries equal to 20 and off-diagonal entries equal to 10. We can express $A$ as: [ A = 10I + 10J ] where $I$ is the $10 \times 10$ identity matrix and $J$ is the $10 \times 10$ matrix of all ones.

The eigenvalues of $J$ are $10$ (with multiplicity 1, corresponding to the eigenvector of all ones) and $0$ (with multiplicity 9, corresponding to vectors with sum zero). Since $A$ is a linear combination of $I$ and $J$, its eigenvalues are given by $\lambda_A = 10 + 10\lambda_J$. Thus, the eigenvalues of $A$ are:

1. For $\lambda_J = 10$: $\lambda_1 = 10 + 10(10) = 110$ (multiplicity 1).
2. For $\lambda_J = 0$: $\lambda_2 = 10 + 10(0) = 10$ (multiplicity 9).

The determinant of $A$ is the product of its eigenvalues: [ \det(A) = 110^1 \times 10^9 = 110 \times 1,000,000,000 = 110,000,000,000 ]

The Python code confirmed this result exactly.

\boxed{110000000000}