111-2004&2005-3-M10-July2005_2

111-2004&2005-3-M10-July2005_2 - Kuwait University...

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Unformatted text preview: Kuwait University Department of Mathematics and Computer science MATH 111, LINEAR ALGEBRA 3 July, 2005, 5—6.15pm. First exam. Time allowed: 75 minutes Answer all questions. Calculators and mobile phones are NOT allowed. 2—~2 L1 1 1 2 _[3_1 1]anng[0 _4 H5]. L/(aj' fpossible, compute BTA, 32 and A + B. ?/ Express the third column of BTA as a linear combination of a (- the columns of BT. VQC) Find the matrix C = [ a b J such that CA : B. c d - U 1 1 r 2. (fipts) Let E 2 0 2 1 , 1 U 1 ' (a) Find det(E). "/ ind all the values of r for which E has an inverse. rzf- 3?: (c) Now suppose 7‘ = 1. Find, if possible, E”. 1 0 U 0 l 2 . . 3. (3pts) Let A 2 1 l k . Flnd the value of k. for wh1ch the 1 2 2}»: . system AX : O has a nontrivial solution. 4. (4131:“ . I _ . Let H : I.” — Bro-w]. where w is an n X m matrix. Show thatfl H is symmetric. } (b) Prrwe that if A is a nonsingular i'i'iatrix, then ;£ 0. l l 1 l 5. (43pt) Answer True 01' False. (Jive reasmis :tor your answer. l d I \ ". 0Q, The product of a lower il‘lé-tllglllal' matrix and a scalar matrix 7'- is lower triangular. lb (b) HA and B are 3x3 matrices with clet(.4) = 4 and det(B) :: 2. hen (lct(A—I(ZB)T) = 1. Q If C- is an 9 X 7 matrix. then CTC' : OCT. O. Math 111, Linear Algebra 3 July. 2005 2 —-l 1. (a) ETA r: [ 7 l0 2 ---5 (1.5111,) [13 111111 possible {if-2111.] :1.11<'i —11 1 —7 3 W1 1 .‘ A+Bfil:3 5 __4](1/2pt). 1 (1)) Third (3011111111 of BTA is — l + —4 (1131;). 2 2 ' 72 7 i (a) 110.4: [ 72:23 :B. we have 2a+3b+ = 1,411— b :1. so (1 = —1.b= 1 a11dQc+3d : (17 e20 i d z 7/1. so (.- : 3.1.1 i -—2. We also notice that —a. + b r: 2. —-C +11. : *5, so CA : B (4pts). 1 1 i" 11 r 2. det(E)=[1"3J1‘1) 0 2 1 :[1'3-1- $11) 0 2 1 :37'21‘ O —11—r‘ Q 0 %_g-! _{2pts). (b) E has an iriverse <:> 1‘ gé (1pt). 1111 U U 1 11 10 (l (c) Whenr=1,[Efg]: 0 210111 2 0 21 0111 = 101001] 07107101 111 10 0 11 1 1 U 0 ' 0 21 0 1 0 = 01 g 0 § 0 00141171 001—212 110 3 —1—2 10 U 2 —1 —1 = 01010—1 = 01010—1,soE1: 0 01 —2 1 2 0 01—2 1 2 —2 1 2 1 0 0 . 0 1 2 _ . . . 3. A is re to D O 2 _ k (2pts), so there W111 be a nontrrvral solution 42> 0 O O k = 2 (lpt). v 4. (a) HT 2 (In - waT)T = If: — (waT)T : In 7 2(wT)TwT : 17n — 211111131, so H is symmetric (2pts). (b) If AA—l : I", then lA||A‘1| = |In| : 1, so |A| # 0 {2pts). 5. (1%pt) each. (a) True. Suppose A is lower triangular and B = r]... a. scalar matrix. Then f01‘f;< we have : (151111;; + ‘ ‘ ‘ + Clijbj‘j + ‘ ' ' + Ginbnj. The terms to the left and right of aijbfi are zero as the row and column indices for B are not equal. The term aijbjj is zero since aij = 0 for i < 3'. Thus AB is lower triangular. (1)) False. det(A) = 4 and det(B) = ‘2, so det(A’1(2B)T) : |A"1||(2B)T| : §|2B|=§2318|22x2=1 . (c) False: CTOiS7X7andCCTiSQX9. fl! .9 ...
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111-2004&amp;amp;2005-3-M10-July2005_2 - Kuwait University...

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