This preview shows pages 1–3. 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: Hour Exam I October 15, 2010 1. Find the inverse of the matrix A = 1 1 1 2 1 2 . Solution: For this you just need to compute the reduced row echelon form of ( A I ): 1 1 1 0 0 1 2 0 1 0 1 2 0 0 1 1 2 0 1 0 1 1 1 0 0 1 2 0 0 1 1 0 2 0 1 0 0 1 1 1 0 0 0 0 1 1 0 1 1 0 0 2 1 2 0 1 0 2 1 0 0 1 1 1 . So A 1 =  2 1 2 2 1 1 1 . 2. Find the solution to x 1 + 2 x 2 + 3 x 3 = 0 2 x 1 + 4 x 2 + 7 x 3 + 2 x 4 = 1 in the form ~x = ~v + s~v 1 + t~v 2 , where s and t are arbitrary real numbers. Solution: Written in augmented matrix form this is 1 2 3 0  2 4 7 2  1 1 2 3 0  0 0 1 2  1 1 2 0 6   3 0 0 1 2  1 That means x 1 x 2 x 3 x 4 =  3 2 s + 6 t s 1 2 t t =  3 1 + s  2 1 + t 6 2 1 2 3. Suppose that A is an invertible n n matrix, B...
View Full
Document
 Fall '08
 lee

Click to edit the document details