{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

homework set 14-Solutions

# homework set 14-Solutions - B ˙ ILKENT UNIVERSITY...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 2-dimensional 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 3-dimensional with basis consisting of the vectors ~v 1 = (1 , 1 , , 0) , ~v 2 = (1 , , 1 ,...
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

homework set 14-Solutions - B ˙ ILKENT UNIVERSITY...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online