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111-2008&amp;2009-3-M10-July2009

111-2008&amp;2009-3-M10-July2009 - Kuwait University...

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Unformatted text preview: Kuwait University Department of h’Iathenlatics and Computer science MATH 111, LINEAR ALGEBRA 14 July, 2009, 6—130an First exam. Answer all questions. Calculators and mobile phones are NOT allowed. —3 2 —1 1. (ths) Let A . Find all constants c E R such that (CAT) - (cA) = 28. ( 2pts) Find —1 [I 2 l (Bpts) Let A—1 that ACBT = {2. J. Find a matrix C such ( 4pts) (a) Show that if A is a 2 \ 3 matrix with AA:r = 02, then A = 02. It —2 ti 5 (1)) Let A = [ Find a 2 x 2 matrix B 7E 02 such that AB = 02. (apt-s) Detern'iinc for which value(s) of k the following sfvstcni has (a) no solution [hi a unique solution (c) inﬁnitclv many solutions: at +29 31: —y 41: +3; "J n. +52 +Ur2 — rap 1. (4px) Let E = O l. C-I—II—l l :3 ,. Find all the values of r for which E has an inverse. Now suppose r = —1. Find, if possible, E‘i. Answer True or False. lGive reasons for your answer. If A is a 3 x 3 skew—symmetric matrix, then det(A) = D. If A and B are 2 x 2 skew—symmetric matrices, then AB is a scalar matrix. If A and B are 3 x 3 matrices with clet(A) = 2 and clet(B) det-(QfA‘llgBT) = —3,a"2. If A is an 3 x 4 matrix, then AX = O has a non—trivial solution. U99 —3_. then AM 1 0 2 3 1 ll '3 3 1 U 2 3 _ II 2. 0 1 _ = U 1 l —L = 1 1 —1 = 14. 0 1 .3 1 0 ll '2 —3 ll 0 2 —3 0 U 2 — U H '2 —1 H U 0 T . . '1 'T 1 '1 . —1-' —1‘-'T _1 1 .3. ALB = 12 E C- =A LB ,1 = 0 9 . LL. Looking 111 1111* ding-111111] _L’;i\'q-:- of] +11% = IIEL +11% =1| :~ .-1= 0-2. 1.! If} —2(? —2f9’ LHF {3H T111923: [1 ” 1| H l 2 —1 l 1 '3 —1 4 13. A11g1111111ted matrix 15 3 —1 5 2 a. 0 —T 11 —10 r —- 121-2—111—1—2 0 —T 1.2—3 L—14 1 2 _:1 __ 0 —T' 1 —1H 0 0 12 — 11: .1- — The 111.51 equation 11mm is 11— 4:112:1'4—112 = I; — :1. Thus [1-1] F: = — 1 {If} 1- aé :4 (e) k = 4. 1 1 1 1 U 0 1 1 1 1 0 0 G[E13]=013l110w01 3010b 1III1II{}1 O —1 r—1 —101 1 0 —2 1 —1 U 0 1 3 0 1 U . T111151 E has inverse 1‘ 7-5 —2 0 0 1+2 —1 1 1 1 U 0 —1 1 2 —1 1 2 For 1“ = —1.~ I] 1 O 3 —2 —.'3 so that 13—1 = 33 —2 —.'3 [I U 1 —1 1 1 —1 1 1 T. (11]: T1110: AT — —A —‘> — If 113|A| — .50 detiA‘j = 0. I. 1. _ __ 0 r1. 0 E1 _ —aE1 0 11),: T1111... [ _ﬂ- 0] [ _b 0] — [ 0 _ab {e} Palm: 110112.443'1’3: \$11—31 = —o. {(1} True: AX = 0 has a 11131141113111 5011111011 1.111102 111111111111“ of variables is . more 1111111 111111111111 of equations. 1 l J ...
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