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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 # 14 U. Mu˘gan July 2, 2008 Homework problems from the 2 nd Edition, SECTION 4.4 3 (3) ) 2 Any four vectors in R 3 are L.D., so the given vectors do not form a basis for R 3 . 8 (8) ) Since det £ ~v 1 ~v 2 ~v 3 ~v 4 / = 66 6 = 0, the given four vectors are L.I. and hence they form a basis for R 4 . 10 (10) ) The single equation y z = 0 is already a system in reduced echelon form, with free variables x and z . With the choice of x = s, y = z = t , we get the solution vector ( x,y,z ) = ( s,t,t ) = s (1 , , 0) + t (0 , 1 , 1) = s~v 1 + t~v 2 . So the given plane is a 2dimensional subspace of R 3 with the basis consisting of the vectors ~v 1 , ~v 2 . 12 (12) ) The typical vector in R 4 of the form ( a,b,c,d ) with a = b + c + d can be written as ~v = ( b + c + d, b, c, d ) = b (1 , 1 , , 0) + c (1 , , 1 , 0) + d (1 , , , 1) . Hence the subspace consisting of all such vectors in 3dimensional with basis consisting of the vectors ~v 1 = (1 , 1 , , 0) , ~v 2 = (1 , , 1 ,...
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 Spring '10
 YosumHoca
 Linear Algebra, Vectors, Vector Space

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