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Unformatted text preview: B ˙ ILKENT UNIVERSITY Department of Mathematics MATH 225, LINEAR ALGEBRA and DIFFERENTIAL EQUATIONS, Solution of Homework set 1 # 15 U. Mu˘gan July 2, 2008 Homework problems from the 2 nd Edition, SECTION 4.5 5) E = 1 0 2 0 0 1 3 0 0 1 . Row basis : The three row vectors of E . Column basis : The first, second, and fourth column vectors of A . 12) Reduced row echelon form of the matrix is E = 1 0 2 1 0 0 1 1 2 0 0 0 0 0 1 0 0 0 0 0 . Row basis : The first three row vectors of E . Column basis : The first, second, and fifth column vectors of A . 14) Consider the matrix A having the given vectors ~v 1 ,~v 2 ,~v 3 ,~v 4 as its column vectors. The reduced row echelon form E of A is E = 1 0 1 / 5 2 0 1 2 / 5 1 0 0 0 0 . Therefore the vectors ~v 1 and ~v 2 are L.I. 16) Similar to the previous problem, the reduced row echelon form of A is E = 1 0 2 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 .....
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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

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