[THOUGHT] We need to compute the determinant of a 10×10 symmetric matrix A. The matrix has a special structure: diagonal entries are 20, off-diagonal entries are 10. So A = 10 * J + 10 * I, where J is the all-ones matrix of size 10×10, and I is the identity matrix. Actually: A = 10J + 10I? Let's check: diagonal entries: 20 = 10 + 10, off-diagonal: 10 = 10 + 0. So A = 10J + 10I? Wait: J has ones everywhere, so 10J gives 10 on all entries. Then adding 10I gives diagonal 20, off-diagonal 10. Yes, A = 10J + 10I.

We need det(A). For n×n matrix of form aJ + bI, eigenvalues are: one eigenvalue = an + b (for eigenvector all-ones), and the rest (n-1) eigenvalues = b (for eigenvectors orthogonal to all-ones). Here a=10, b=10, n=10. So eigenvalues: λ1 = 1010 + 10 = 110, λ2...λ10 = 10. Determinant = product of eigenvalues = 110 * 10^9 = 110 * 1,000,000,000 = 110,000,000,000.

But we should verify with Python to avoid arithmetic errors.

Let's compute determinant directly using numpy or sympy. </think>

[PYTHON] import numpy as np

Construct matrix A

n = 10 A = np.full((n, n), 10) # all entries 10 np.fill_diagonal(A, 20) # set diagonal to 20

Compute determinant

det = np.linalg.det(A) print("Determinant (float):", det)

Since determinant is an integer, we can round to nearest integer

det_int = int(round(det)) print("Determinant (integer):", det_int)

Alternatively, compute using eigenvalues

eigvals = np.linalg.eigvals(A) print("Eigenvalues:", eigvals) print("Product of eigenvalues:", np.prod(eigvals)) print("Product (rounded):", int(round(np.prod(eigvals)))) [/PYTHON]

[THOUGHT] The Python output shows the determinant as approximately (1.1 \times 10^{11}), specifically 110000000000 when rounded to the nearest integer. The eigenvalues confirm this: one eigenvalue is 110 and the rest are 10 (with negligible numerical errors). The product of eigenvalues is (110 \times 10^9 = 110000000000). This matches the integer determinant.

[/THOUGHT] \boxed{110000000000}