Assignment5Solution(1804)

Assignment5Solution(1804) - MATH1804/AS5sol/ZJH/2009 THE...

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

View Full Document Right Arrow Icon
MATH1804/AS5sol/ZJH/2009 THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH1804: University Mathematics A Solution to Assignment 5 1. (a) not well-defined (b) not well-defined (c) BA = - 1 6 1 0 ! 1 2 4 0 1 - 1 ! = - 1 4 - 10 1 2 4 ! (d) AC = 1 2 4 0 1 - 1 ! 2 0 - 1 1 1 2 = 4 10 - 2 - 1 ! (e) not well-defined (f) CB 2 = 2 0 - 1 1 1 2 - 1 6 1 0 ! 2 = 14 - 12 - 8 12 5 6 (g) A T A = 1 0 2 1 4 - 1 1 2 4 0 1 - 1 ! = 1 2 4 2 5 7 4 7 17 (h) 2 B T AC = 2 - 1 6 1 0 ! T 4 10 - 2 - 1 ! = 2 - 6 - 11 24 60 ! = - 12 - 22 48 120 ! 2. (a) Let A = 1 1 - 1 - 1 ! 6 = 0 2 × 2 , but we have A 2 = 1 1 - 1 - 1 ! 1 1 - 1 - 1 ! = 0 2 × 2 (b) Let A = 1 0 0 - 1 ! 6 = I 2 , but we have A 2 = 1 0 0 - 1 ! 1 0 0 - 1 ! = I 2 . 3. Let A = ( a ij ) n × n . By the condition A = 3 A T , we have ( a ij = 3 a ji a ji = 3 a ij which implies that a ij = 0 for any i and j. Thus, A must be the zero matrix. 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
Alternative method: A = 3 A T = 3(3 A T ) T = 9( A T ) T = 9 A . Hence A must be the zero matrix. 4. det A = a ( ei - fh ) - b ( di - fg ) + c ( dh - eg ) while det A T = det a d g b e h c f i = a ( ei - fh ) - b ( di - fg
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.

This note was uploaded on 10/14/2010 for the course MATH math1804 taught by Professor Prof during the Spring '08 term at HKU.

Page1 / 6

Assignment5Solution(1804) - MATH1804/AS5sol/ZJH/2009 THE...

This preview shows document pages 1 - 3. Sign up to view the full document.

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