Question

A magic matrix. Consider the nxn matrix Mn obtained by filling the rows of this matrix with the numbers 1, 2,..., n², so that the first row consists of the numbers 1, 2,...,n, the second row of the numbers n+1, n + 2,..., 2n, and so on. Now choose n entries in this matrix such that each row and each column contains exactly one of these entries, and add these n entries. Prove that, no matter how these n entries are chosen, for a given n-value their sum is always the same, and equal to a certain number Sn, and find a formula for Sn.

Answer 1

To prove that the sum of the chosen entries in the magic matrix is always the same for a given n, we can observe a pattern and use series expansion. The formula for the sum Sn is Sn = 2n².

Step-by-step explanation:

To prove that the sum of the chosen entries in the magic matrix is always the same for a given n, we can use the concept of series expansion. Let's consider the matrix Mn, where each row consists of consecutive numbers from 1 to n, n+1 to 2n, and so on. Now, choose n entries in this matrix such that each row and each column contains exactly one of these entries. The sum of these entries can be represented as Sn.

We can observe that by selecting (n-1) from the last term and adding it to the first term of each row, the resulting sum is 2[n+3+...(2n-3)+n]. Similarly, by selecting (n-3) from the penultimate term and adding it to the second term of each row, the resulting sum is 2n². This pattern continues, and we can prove that the sum of the chosen entries is always 2n².

Therefore, the formula for Sn would be Sn = 2n².