solution9 - M341 Homework Solution 9 Section 4.3: 1(b): We...

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

View Full Document Right Arrow Icon
M341 Homework Solution 9 Section 4.3: 1(b) : We form the matrix whose rows are the vectors in S and row reduce. A = 3 1 2 3 1 2 6 2 4 3 1 2 0 0 0 0 0 0 = C Since Row ( A ) =Row ( C ) =Span S . This tells us that Span S = { a [3 , 1 , 2] : a R } 2(b) : We use the correspondence between R 4 and P 3 . We let ˆ S = { [1 , 2 , 0 , 0] , [0 , 4 , 0 , 1] , [ 5 , 0 , 0 , 12] , [1 , 1 , 0 , 0] } We form the matrix whose rows are the vectors in ˆ S . We form the matrix whose rows are the vectors in ˆ S and row reduce. A = 1 2 0 0 0 4 0 1 5 0 0 12 1 1 0 0 1 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 = C This tells us that Span ˆ S = { a [1 , 0 , 0 , 0]+ b [0 , 1 , 0 , 0]+ c [0 , 0 , 0 , 1] : a,b,c R } = { [ a,b, 0 ,c ] : a,b,c R } . Correspondingly Span S = { ax 3 + bx 2 + c : a,b,c R } . 3(b) : We use the correspondence between M 22 and R 4 . We let ˆ S = { [1 , 3 , 2 , 1] , [ 2 , 5 , 3 , 1] , [1 , 4 , 3 , 4] } We form the matrix whose rows are the vectors in ˆ S . We form the matrix whose rows are the vectors in ˆ S and row reduce. A = 1 3 2 1 2 5 3 1 1 4 3 4 1 0 1 8 0 1 1 3 0 0 0 0 = C This tells us that Span ˆ S = { a [1 , 0 , 1 , 8] + b [0 , 1 , 1 , 3] : a,b R } . Correspondingly Span S = { a [ 1 0 1 8 ] + b [ 0 1 1 3 ] : a,b R } = { [ a b a b 8 a + 3 b ] : a,b R } . 6
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/05/2011 for the course M 341 taught by Professor Hietmann during the Spring '08 term at University of Texas at Austin.

Page1 / 4

solution9 - M341 Homework Solution 9 Section 4.3: 1(b): We...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online