Let A_{n} the nn matrix such that a i, j =2 delta i,j - delta i+1,j - delta i,j+1 where delta t, j equals 1 if i = j and zero otherwise. Compute det(An).

To solve for the determinant of the matrix \( A_n \), where \( A_n \) is defined as \( (a_{i,j}) \) with \( a_{i,j} = 2 \delta_{i,j} - \delta_{i+1,j} - \delta_{i,j+1} \), we need to interpret the given components correctly. Here, \( \delta_{i,j} \) is the Kronecker delta, which is 1 if \( i = j \) and 0 otherwise.

Let's start by constructing the matrix \( A_n \) explicitly for a general \( n \). The matrix \( A_n \) is tridiagonal with the following structure:

- The main diagonal entries are all 2.
- The subdiagonal and superdiagonal entries are all -1.

For example, for \( n = 3 \), the matrix \( A_3 \) would be:

\[
A_3 = \begin{pmatrix}
2 & -1 & 0 \\
-1 & 2 & -1 \\
0 & -1 & 2
\end{pmatrix}
\]

In general, for any \( n \), the matrix \( A_n \) looks like:

\[
A_n = \

Learn how to compute the determinant of matrix A_n, defined as a tridiagonal matrix with specific entries involving Kronecker deltas. This problem explores matrix algebra, determinants, and the properties of tridiagonal matrices. Suitable for advanced mathematics enthusiasts and students.

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