This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Math 419 Solutions - Sections 4.1-4.3 Winter 2010 26 from 4.2 It is easy to see that the first column the matrix A is a 1 = 1 2 3 . . . n Similarly, the second column is a 2 = 2 a 1 ; the third column is a + 3 = 3 a 1 ; . . . ; the n th column is a n = na 1 . These are l.d., hence det A = 0. (if n = 1, then det A = 1). 16 from 4.3 (a) Consider the first three columns of A . Note that the system c 1 a 1 + c 2 a 2 + c 3 a 3 = 0 is equivalent to a system of two equations with three variables. Thus it has to have a nontrivial solution, and so the first three columns of A are l.d.. Hence the columns of A are l.d., and so det( A ) = 0. (b) In the big formula (page 212) we see that det( A ) is a sum of 120 numbers, each of these is a product of five numbers. These products are made of five numbers, one from each column and row. The precise meaning of one from each column and row is that once you pick a number from one column the next number comes from a different row in...
View Full Document
- Spring '08
- Signal Processing