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111-2005&amp;2006-3-M10-June2006_2

# 111-2005&amp;2006-3-M10-June2006_2 - Kuwait University...

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Unformatted text preview: Kuwait University Math. 111 June 26, 2006 Dept. of Math. 85 Comp, Sc. First Midterm Exam Duration 90 Minutes ——_%__—m___._____m—______m Calculators and mobile phones are not allowed Answer the following questions. 1 2 1.LetA: 3 —1 andB=[ 1—2 3] #1 3 31—2 “(an [2 Points] If possible, compute A + B and (MT ~— B)? //(b) [2 Points] If possible, compute AB. 1 3 1 —1 -1 3 2. LetA= 2 O —1 andB: 2 3 m6 1 —1 O 1 l ——2 1/3) [3 Points] Find B-l. , (Vb) [2 Points] Find the matrix C such that A‘lBTC : I3. ﬂ [4 Points] Solve the linear system y + 23 + w = —3 x + 2y — z + 21.1; : 5 22: — y + z — w : ~l f4. [5 Points] Find all values of a and b for which the resulting system has: (i) no solution, (ii) inﬁnitely many solutions, (iii) a unique solution. :2: + y -+~ 32 = 2 a: + 2y + 42 : 3 :5 + 31; + oz = b. a1 :12 a3 0.1 51 61 K 5. [3 POlIltS] If 0.2 52 02 = 3, ﬁnd —2bl —2b2 —2b3 03 53 Cs 30.1 — Cl 30,2 — C2 303 "- C3 9/" 6. Ja) [2 Points] prove that if A is skew symmetric and nonsingular matrix, then A"1 is skew symmetric matrix. /(b) [2 Point] Show that if A and B are nonsingular matrices, then AB is nonsingular and (AB)—1 =B‘1A‘1 0w" mil Kuwait University Math. 111 June 26, 2006 Dept. of Math. Sc Comp. Sc. First Midterm Exam Solution l. a (A + B) is not possible (2ATuB)T=2A—BT=2 if u 112 3 T: 21413 3 1—2 —13 ~58 12 70~1 1.101113: 3—1 [g ‘f j]: 0—7 11 -13 85~9 2.3 —1~13..100 11—35—100 [3513]: 23—0 010~0 0 210 11—2 001 001 101 1013 41—10 10050—13 ~0105210~0105210—[I358‘1] 001 101 0015101 ' 0 “1 3 T 1 3 5 —1 ,2 2. bC:(BT)"1A=(B-1)TA_ 2 1 0 2 0 — : 1 —3 —2 1 0 1 1 —l 0 4 8 3 3. U 1 2 1 . —3 I 2 ——1 2 . 5 [A33] = 1 2 ‘1 2 3 5 N O 1 2 1 5 —3 2 —l 1 —1 f —1 2 -—1 1 m1 5 —~1 1 2 —1 2 E 5 1 O —5 O 3 11 N 0 1 2 1 E —3 N 0 1 2 1 E -3 0 —5 3 —5 3 ~11 0 0 13 0 5 —26 1 0 U 0 i 1 N 0 1 0 1 E 1 0 O I 0 3 —2 There are inﬁnetely many solutions :1: =1, 3; =1— 1*, z = —2, w : 1"; 1” E R. 1 “‘5’” 113:2 113.2 [AEB]=12433~01131 13a‘b 02a—3 b—2 102.1 ~01131 0 0 (1—— 5 E b— 4 a = 5 and b 75 4 => No solution a = 5 and b = 4 => Inﬁnitely many solutions a % 5 => Unique solution (11 b1 C1 61 a2 03 a1 02 (13 (12 b2 62 =3:> b1 b2 [)3 =3 :> —2b1 —-2b2 —2b3 :6 (13 b3 Ca 01 C2 Ca —01 —C2 —C3 (11 (12 as :- —2b1 —2b2 —2b3 =6 3a1 -— 61 3112 — c2 3(13 — C3 6. a A is skew symmetric => AT 2 —A (A'1)T = (AT)_1 2 (—A)‘1 = — (A‘l) => A—1 is skew symmetric 6. b Theorem 1.10, Part b +— g/W wag) “" .2 13",“ A 2 WWW): quirsﬁa’l :— ,4 :1, A ‘ nae/2:1,” (lg/61y 74 75) :2 BYA/ 143.3,. [ff/71g :B'lg :3” OO 99 ...
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