This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: B ˙ ILKENT UNIVERSITY Department of Mathematics MATH 225, LINEAR ALGEBRA and DIFFERENTIAL EQUATIONS, Solution of Homework set 1 # 6 U. Mu˘gan June 20, 2008 Homework problems from the 2 nd Edition, SECTION 3.2 3 (3) ) 2 Since the linear system is already in echelon form, we can find the solution of the system by using the back substitution. If we set x 3 = t then the 2 nd eq. gives x 2 = 2 + 5 t and the 1 st eq. gives x 1 = 13 + 11 t . 6 (6) ) Since the linear system is already in echelon form, we can find the solution of the system by using the back substitution. If we set x 3 = t and x 4 = 4 from the 3 rd eq., then the 2 nd eq. gives x 2 = 11 + 3 t and the next eq. gives x 1 = 17 + t . 9 (9) ) Since the linear system is already in echelon form, we can find the solution of the system by using the back substitution. Start with x 4 = 6 from the 4 th eq., the 3 rd eq. gives x 3 = 5, the 2 nd eq. gives x 2 = 3, and finally the 1 st eq. gives x 1 = 1....
View
Full
Document
This note was uploaded on 06/18/2010 for the course COMPUTER S Math 225 taught by Professor Yosumhoca during the Spring '10 term at Bilkent University.
 Spring '10
 YosumHoca

Click to edit the document details