ยูฮู

ยูฮู

ผลลัพธ์ (อังกฤษ) 1:

Yu Hu

ผลลัพธ์ (อังกฤษ) 2:

Yoo

ผลลัพธ์ (อังกฤษ) 3:

Chapter 3 SOLUTIONS TO EXERCISES Exercise 3.1. Prove or disprove: (a) tr (AB) = tr (A) tr (B). (b) tr (AB) = tr (BA). (c) tr (A. + B) = tr (A) + tr (B). (d) tr (A − 1) = 1 tr (A). (E) tr (xA) = x tr (A). (Here x is a scalar.) (f) det (xA) = x det (A). (g) det (A. + B) = det (A) + det (B). (H) det (xI − A) = x 2 − tr (A) x + det (A). Answers: a) False. For example if A = I and B is any matrix. With tr (B) 6 = 0 we have tr (AB) = tr (B), while tr (A) tr (B) = 2 tr (B) 6 = tr (B). B) True. Tr (a B e d x y z w) = ax + BZ. + +, cy DW which is unchanged if you switch the matrices. C) True. Tr (a B e d + X Y Z w) = a+x+d+w which is, unchanged. If you switch the matrices. D), False even for the identity matrix. E) True. Tr (x a B C D) = XA + XD = x (a + D) = x tr. A B C D). F), False even for the identity matrix. What is true: det (xA) = x 2 det (A). G) False. Take for example A =,,, I B = −, I. H) True. Just compute it.