Consider the 10$$\times$$ 10 real symmetric matrix
Calculate
[THOUGHT] The given matrix A is a 10x10 symmetric matrix with diagonal entries 20 and off-diagonal entries 10. This matrix can be expressed as ( A = 10I + 10J ), where ( I ) is the identity matrix and ( J ) is the matrix of all ones. The determinant of such a matrix is given by the formula ( \det(A) = (a + nb) \cdot a^{n-1} ), where ( a = 10 ), ( b = 10 ), and ( n = 10 ). Applying the formula:
[ \det(A) = (10 + 10 \cdot 10) \cdot 10^{10-1} = 110 \cdot 10^9 = 110000000000 ]
To verify, we will construct the matrix and compute its determinant using Python. [/THOUGHT] [PYTHON] import numpy as np
n = 10 A = np.full((n, n), 10) np.fill_diagonal(A, 20) det = np.linalg.det(A) print(round(det)) # Rounding to handle floating-point precision issues [/PYTHON]
[THOUGHT] The Python code output confirms the determinant of matrix ( A ) is ( 110000000000 ). This matches the analytical formula for matrices of the form ( aI + bJ ), where ( a = 10 ), ( b = 10 ), and ( n = 10 ). The result is consistent and error-free. [/THOUGHT] \boxed{110000000000}