Difference between revisions of "2011 AIME II Problems/Problem 11"
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This is an infinite geometric sequence with first term <math>\frac{1}{81}</math> and common ratio <math>\frac{1}{9}</math>. Thus, the sum is <math>\frac{\frac{1}{81}}{1-\frac{1}{9}}=\frac{\frac{1}{81}}{\frac{8}{9}}=\frac{9}{(81)(8)}=\frac{1}{(9)(8)}=\frac{1}{72}</math>. | This is an infinite geometric sequence with first term <math>\frac{1}{81}</math> and common ratio <math>\frac{1}{9}</math>. Thus, the sum is <math>\frac{\frac{1}{81}}{1-\frac{1}{9}}=\frac{\frac{1}{81}}{\frac{8}{9}}=\frac{9}{(81)(8)}=\frac{1}{(9)(8)}=\frac{1}{72}</math>. | ||
− | Thus, p + q = 1 + 72 = <math>\framebox[1.1\width]{ | + | Thus, p + q = 1 + 72 = <math>\framebox[1.1\width]{073.}</math> |
Revision as of 09:55, 1 April 2011
Problem:
Let be the n x n matrix with entries as follows: for
,
; for
,
; all other entries in
are zero. Let
be the determinant of the matrix
. Then
can be represented as
where p and q are relatively prime positive integers. Find p + q.
Note: The determinant of the 1 x 1 matrix is a, and the determinant of the 2 x 2 matrix
is ad-bc; for n
2, the determinant of an n x n matrix with first row or first column
. . . .
is equal to
where
is the determinant of the (n-1) x (n-1) matrix found by eliminating the row and column containing
.
Solution:
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 sequence with first term and common ratio
. Thus, the sum is
.
Thus, p + q = 1 + 72 =