Final answer:
The given matrix A is symmetric because the elements are arranged in such a way that element at position (i, j) is the same as the one at position (j, i) for all i and j, fulfilling the condition for a matrix to be symmetric.
Step-by-step explanation:
To determine whether a given matrix A is symmetric, we need to check if the element at position (i, j) is the same as the one at position (j, i) for all i and j. In the matrix A provided, where elements are defined as: 1+a_1, a_2, a_3, ... a_n on the first row, a_1, 1+a_2, a_3, ... a_n on the second row, etc., we observe a pattern. The diagonal elements are of the form 1+a_i, where i depends on the row/column number, while the off-diagonal elements are a_n, with the position in the row being irrelevant.
For a matrix to be symmetric, a_ij must be equal to a_ji. That is, the element in the ith row and jth column should be identical to the element in the jth row and ith column. Here, the off-diagonal elements are consistent as they do not depend on the row/column order, merely on their position within the sequence. As a result, this matrix is symmetric since for any element a_ij = a_ji.