This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Therefore we get A = E1 1 E1 2 E1 3 (or A = E1 1 E1 2 E1 3 I ) as a solution where E1 1 = 0 1 0 1 0 0 0 0 1 , E1 2 = 1 1 0 0 1 0 0 0 1 , E1 3 = 1 0 0 0 1 0 4 0 1 . (b) We have det( A ) =det( E1 1 E1 2 E1 3 I ). Here E1 2 and E1 3 are type III and E1 1 is type I. Since type III operations do change the determinants while type I change the signs, we get det( A ) =det( I ) =1....
View Full
Document
 Spring '08
 KIM
 Linear Algebra, Algebra, Matrices

Click to edit the document details