[THOUGHT] We need to compute the determinant of a 10×10 real symmetric matrix A. The matrix has 20 on the diagonal and 10 elsewhere. This is a special matrix: A = 10J + 10I, where J is the all-ones matrix, and I is the identity matrix? Let's check: A_{ii} = 20 = 10 + 10, so A = 10J + 10I, where J is matrix with all entries 1. Indeed, 10J gives 10 everywhere, plus 10I adds 10 on diagonal, total diagonal 20, off-diagonal 10. So A = 10*(J + I). Then det(A) = det(10*(J + I)) = 10^10 * det(J + I). So we need det(J + I). J is rank 1, eigenvalues: one eigenvalue = n (since Jones = nones), others zero. J + I shifts eigenvalues by 1: eigenvalues of J+I are: (n+1) once, and 1 repeated (n-1) times. For n=10, eigenvalues: 11 (once) and 1 (nine times). So det(J+I) = 11 * 1^9 = 11. Then det(A) = 10^10 * 11 = 11 * 10^10 = 110000000000. But let's verify with Python to be sure.

We can compute directly using numpy.linalg.det or sympy. However, for large numbers, floating point may give scientific notation. We'll use sympy for exact integer determinant.

Approach: construct matrix A as sympy Matrix, compute det.

Let's do it. [/THOUGHT] [PYTHON] import sympy as sp

n = 10

Create matrix A: diagonal 20, off-diagonal 10

A = sp.Matrix([[20 if i==j else 10 for j in range(n)] for i in range(n)]) detA = A.det() print(detA) [/PYTHON]

[THOUGHT] The Python output shows the determinant is 110000000000. This matches the eigenvalue reasoning: det(A) = 11 * 10^10 = 110000000000. So the answer is 110000000000. [/THOUGHT] \boxed{110000000000}