{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw9s - MATH 4377 ADVANCED LINEAR ALGEBRA SUMMER 2011 Key to...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
MATH 4377, ADVANCED LINEAR ALGEBRA, SUMMER 2011, Key to HW#9 Note : All the problems below are related to Theorem 3.16 on Page 191. So please read the understand Theorem 3.16 before doing the following home- work. 1. Problem 5 Page 196. Solution . By Theorem 3.16, the third column of A is 2 1 - 1 3 - 5 0 - 1 1 = 2 3 1 . The fifth column of A is - 2 1 - 1 3 - 3 0 - 1 1 + 6 1 - 2 0 = 4 - 7 - 9 . Hence A = 1 0 2 1 4 - 1 - 1 3 - 2 - 7 3 1 1 0 - 9 . 2. Problem 7 on Page 197. Solution . 2 1 - 8 1 - 3 - 3 4 12 37 - 5 1 - 2 - 4 - 17 8 -→ 1 - 2 - 4 - 17 8 0 5 0 35 - 19 0 - 2 0 14 - 19 -→ 1 0 - 4 - 31 27 0 1 0 - 7 19 / 2 0 0 0 70 - 133 / 2 -→ 1 0 - 4 0 * 0 1 0 0 * 0 0 0 1 - 133 / 140 So { u 1 , u 2 , u 4 } is a basis for R 3 . 1
Background image of page 1

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

View Full Document Right Arrow Icon
3. Problem 9 on Page 197. Solution . 0 1 2 1 - 1 - 1 2 1 - 2 2 1 3 9 4 - 1 -→ 1 3 9 4 - 1 0 1 2 1 - 1 0 0 0 - 3 6 So { u 1 , u 2 , u 4 } is a basis for W. 4. Problem 12 on Page 197. Solution . 12(a) It is easily verified that S is a linearly independent subset of V .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}