Question
21/le/content/352479/viewContent/3310368/View?ou = 352479 1020 1050 1040 St. John's, NL - 7 Da... MEMORIAL UNIVERSITY OF NEWFOUNDLAND...
21/le/content/352479/viewContent/3310368/View?ou = 352479 1020 1050 1040 St. John's, NL - 7 Da... MEMORIAL UNIVERSITY OF NEWFOUNDLAND DEPARTMENT OF MATHEMATICS AND STATISTICS © Solomon Assignment 8 Mathematics 2050 Spring 2020 Due: July 30, 2020, 11:59 pm . SHOW ALL WORK a b c -T (3) 1. If pg T -1, compute 3p+a 3q+b3r + 2p 24 2r [2] [2] 2. (a) What are the possible values of det(A) if A-' = A' ? (b) A matrix is skew symmetric if A" = -A. If A is a skew - symmetric matrix of odd order, show that A is not invertible. (c) Let A and B ben x n matrices with det(A) = det(B) +0. If a and b are non-zero real numbers, show that det(aA +bB-1) = det(aB+bA-) [2] 3. Use determinants to find which real values of c make the matrix с 0 -1 1 0 2 с invertible. c [12] 4. Find the characteristic polynomial, the eigenvalues and corresponding eigenvectors of each of the following matrices 1 2 - 2 (a) 3 2 (b) 6 -6 2
Answers
0 Here, a P be qin Now, -2 3 pta 39th 3rtel 2 + 29 2Y - -Z 10 2 apta 39th arte Р q r - Z -9 2/2 3pt a 34+ b 3P 39 3Y+C -n - y z win [22's R2-R3] c 3P 32 3r X -Y 1) N|دی 3 a b q - 2 1 که کا a JE P q P = -2:6-1 q b es i Exchanging Rin and a С. n y ㅋ -2.(-1) (-1) Р n مجله Exchanging Rp and md Roz y 2 -2. (-1)(-1).(-1) (Any). - 2@ detero, ATAT det (A) =ded (17)] -> A AT = 1 det CAAT) = d'ed (F) > det (A) det lat) 5) deten) detCA) = 1 [ (ded (A)) ² det (A) = 1 2 Et Her A АТ > det (AT) = del GA) → det (A) = -detta) [": Jel (n) =detit] >> 2ded (A) deti) A is not invertible. 0 SO Here, det lA) = d-est LA) 0. - Now, det (aA tb B-1) りサ = det (a A BB"'+bA A-18-1) [: AN-'=64-1 dett (25+ ba-) 8-] det (A) del (ab tb A") der (B-1) = det in) sed (aB+64-1(dlet Ls)) det (ab to A-1) [- det (A) = ded CBD det (aA +68-1) = det caB rb A-1)(3) consider, o det 0 2 С. 1 с. 7 (2-3)+(-)=0 o(2-c²_-) so (+-(2) o = C-0, +1,-1. . the above matrix ip invertible if C #0, c1, C F - (Any).
![@ detero, ATAT det (A) =ded (17)] -> A AT = 1 det CAAT) = ded (F) > det (A) det lat) 5) deten) detCA) = 1 [ (ded (A)) ² det](https://i.imgur.com/996T8qO.png)
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