2011 AIME II Problems/Problem 11
Let be the matrix with entries as follows: for , ; for , ; all other entries in are zero. Let be the determinant of matrix . Then can be represented as , where and are relatively prime positive integers. Find .
Note: The determinant of the matrix is , and the determinant of the matrix ; for , the determinant of an matrix with first row or first column is equal to , where is the determinant of the matrix formed by eliminating the row and column containing .
Using the expansionary/recursive definition of determinants (also stated in the problem):
This pattern repeats because the first element in the first row of is always 10, the second element is always 3, and the rest are always 0. The ten element directly expands to . The three element expands to 3 times the determinant of the the matrix formed from omitting the second column and first row from the original matrix. Call this matrix . has a first column entirely of zeros except for the first element, which is a three. A property of matrices is that the determinant can be expanded over the rows instead of the columns (still using the recursive definition as given in the problem), and the determinant found will still be the same. Thus, expanding over this first column yields . Thus, the expression turns into . Thus, the equation holds for all n > 2.
This equation can be rewritten as . This version of the equation involves the difference of successive terms of a recursive sequence. Calculating backwards from the recursive formula and from the formula yields . Examining the differences between successive terms, a pattern emerges. , , , , . Thus, .
Thus, the desired sum is
This is an infinite geometric series with first term and common ratio . Thus, the sum is .
We calculate the first terms of . They are: and . From here, we can tell that it is just ~Lcz
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