<think> Okay, I need to calculate the determinant of this 10x10 symmetric matrix A. Let me first understand what the matrix looks like. Each diagonal element is 20, and all off-diagonal elements are 10. So it's a matrix where every entry is 10 except for the diagonals which are 20.

Hmm, calculating the determinant of such a large matrix directly seems complicated. Maybe there's a pattern or property I can use to simplify this. Since A is symmetric, perhaps it has some special structure that makes finding its determinant easier.

Wait, all rows and columns have similar structures except for the diagonal. This reminds me of matrices where each row sums to the same value. Let me check: each row has 20 on the diagonal and nine 10s off-diagonal. So the sum of each row is 20 + 9*10 = 20 + 90 = 110.

If all rows sum to the same value, that suggests that the vector with all ones might be an eigenvector. Let me think about eigenvalues because if I can find them, the determinant would just be the product of the eigenvalues.

So suppose v is a vector where each component is 1. Then Av should be equal to λv for some eigenvalue λ. Calculating Av: Each entry in Av will be 201 + 910*1 = 20 + 90 = 110, so Av = 110v. Therefore, one eigenvalue is 110.

Now, what about the other eigenvalues? Since A is a symmetric matrix, it's diagonalizable, and all its eigenvalues are real. The trace of A is the sum of the diagonal elements, which is 10*20 = 200. But since one eigenvalue is 110, the sum of the remaining nine eigenvalues must be 200 - 110 = 90.

Wait, but I need more information about the other eigenvalues. Let me consider that A can be written as a rank-one matrix plus a diagonal matrix. Specifically, A = D + 10*(ee^T), where e is a vector of ones and D is a diagonal matrix with entries 20 - 10 = 10 on the diagonal.

So D is a 10x10 matrix with 10s on the diagonal, and ee^T is a matrix where every entry is 1. Therefore, A can be expressed as:

A = 10I + 10ee^T

Where I is the identity matrix. Now, this form might help in finding eigenvalues.

I remember that if you have a matrix of the form aI + b ee^T, then its eigenvalues are a + b*n (where n is the size) and a with multiplicity n-1. Let me verify that.

In our case, a = 10 and b = 10, but wait, actually A = 10I + 10ee^T. So the eigenvalues would be:

For the vector e: (10I)e + 10(ee^T)e = 10e + 10*(sum of e components)e. Since sum(e) is 10, this becomes 10e + 1010*e = 110e. So the eigenvalue corresponding to e is 110.

For vectors orthogonal to e: Let w be a vector such that w^T e = 0. Then A w = (10I + 10 ee^T)w = 10w + 10(e^Tw)e. But since w is orthogonal to e, e^Tw = 0, so Aw = 10w. Therefore, the other eigenvalues are all 10.

So in total, we have one eigenvalue of 110 and nine eigenvalues of 10 each.

Therefore, the determinant of A is the product of its eigenvalues: 110 * (10)^9.

Let me compute that. 10^9 is 1,000,000,000. So 110 multiplied by that is 110,000,000,000.

Wait, let me double-check the eigenvalues to make sure I didn't make a mistake. The matrix A can be written as 10*(I + ee^T). Wait no, because each off-diagonal element is 10, so actually, it's D + 10*ee^T where D has 20 on the diagonal and 10 elsewhere? Hmm, maybe I need to adjust my earlier reasoning.

Wait, let me clarify: The original matrix A has 20 on the diagonal and 10 elsewhere. So we can write A as 10*(ones_matrix) + (20 - 10)I = 10ones_matrix + 10I. Because ones_matrix has all entries equal to 1, so multiplying by 10 gives a matrix with 10s everywhere except the diagonal which would be 10 as well. But in A, the diagonal is 20, so we need to add another 10 on the diagonal, hence 10I.

So yes, A = 10*(ones_matrix) + 10*I.

Now, the eigenvalues of ones_matrix are known: it has rank 1, so one eigenvalue is n (which is 10 here), and the rest are zero. Therefore, when we multiply by 10, the eigenvalues become 10*10=100 for the vector e, and 0 otherwise.

Then adding 10*I to this matrix shifts all eigenvalues by 10. So the eigenvalue corresponding to e becomes 100 + 10 = 110, and the other eigenvalues become 0 + 10 = 10 each.

So that confirms my earlier conclusion: one eigenvalue of 110 and nine eigenvalues of 10.

Thus, determinant N is 110 * (10)^9. Calculating that gives 110 followed by 9 zeros, which is 110,000,000,000. </think>

[THOUGHT] To calculate the determinant of matrix A, we recognize its structure as a symmetric matrix with diagonal entries 20 and off-diagonal entries 10. By expressing A in terms of rank-one matrices and leveraging eigenvalue properties, we determine that A has one eigenvalue of 110 and nine eigenvalues of 10 each. The determinant is the product of these eigenvalues. [/THOUGHT] [PYTHON]

Calculate the determinant as the product of eigenvalues

eigenvalue_1 = 110 eigenvalue_others = 10 ** 9 determinant = eigenvalue_1 * eigenvalue_others print(determinant)

<think> Alright, so I'm trying to figure out how to calculate the determinant of this 10x10 matrix A. The matrix has a specific structure: each diagonal element is 20, and every off-diagonal element is 10. That's interesting because it looks like all the rows are very similar except for one element which is higher.

First, I remember that determinants can be tricky to compute directly, especially for large matrices like this one. There must be a pattern or property we can exploit here. Since the matrix is symmetric and has a lot of repetition in its entries, maybe there's a way to simplify it without having to perform row operations or expand cofactors manually.

I recall that if a matrix can be expressed as a sum of simpler matrices, especially rank-one updates, it might have eigenvalues that are easier to compute. In this case, the matrix A seems like it could be written in terms of a scalar multiple of the identity matrix plus another matrix with all entries equal to 10.

Let me think: If I let J be a 10x10 matrix where every entry is 1, then perhaps A can be expressed as 20I + 10(J - I). Wait, that simplifies to (20 - 10)I + 10J = 10I + 10J. So A = 10(I + J).

Now, the determinant of a matrix scaled by a scalar is the scalar raised to the power of the size of the matrix times the determinant of the original matrix. But in this case, since I'm dealing with a sum of matrices, it's not straightforward.

However, I remember that if you have a rank-one update to a diagonal matrix, there's a formula for its eigenvalues. The matrix J is a rank-one matrix because all its rows are the same. So, the matrix I + J has a specific set of eigenvalues. Specifically, since J has rank 1, it has one non-zero eigenvalue equal to the trace of J, which is 10 (since each diagonal entry is 1 and there are 10 entries). The other eigenvalues are zero.

Therefore, the matrix I + J will have eigenvalues: 1 + 10 = 11 for the rank-one part and 1 for the remaining nine dimensions. So, the eigenvalues of I + J are 11 (once) and 1 (nine times).

Now, since A is equal to 10(I + J), the eigenvalues of A will be 10 times the eigenvalues of I + J. That means the eigenvalues of A are 10 * 11 = 110 (once) and 10 * 1 = 10 (nine times).

The determinant of a matrix is the product of its eigenvalues. So, the determinant N of A will be 110 multiplied by 10 raised to the ninth power.

Calculating that: 110 * (10^9) = 110,000,000,000.

So, the final answer is 110 followed by nine zeros. </think>

The determinant of matrix A is calculated as the product of its eigenvalues. Given that one eigenvalue is 110 and the remaining nine are each 10, the determinant N is:

[ N = 110 \times 10^9 = 110,000,000,000 ]

Thus, the final answer is:

\boxed{110000000000}