111-2007&2008-2-M10-March2008_2

111-2007&2008-2-M10-March2008_2 - Kuwait University...

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Unformatted text preview: Kuwait University MATH 111 Mar 31, 2008 Math. Dept. First Midterm Test Duration 90 minutes Calculators and all mobile phones are not allowed during the exam. L [2pts.] LetAbeaZXEmstrixsndfiz _? [fAB= i]! find A. 2. [2 pts.] Prove that if A and B are both 11 3< n skew—mnmletric matrices. then AB + 3.4 is synunetric. 132 3—13 &[3pm[I£L4= 4 0 1 ,amiB_1= 1 1 —2 . 7 —1 5 —2 —2 —2 0 Find a. 3 K 3 matrix C. such that 31-1—10 = [3. 4. [3+2 pts.[ :1. Show that if A is non—singular, then 3E 0. h. Prove or disprove: If .4 and B are non—singular n )< 1*: matrices, then A + B is non—singular. 5. [3 pts.] Let = 4 and [3—1[ = —2 where A. and B are both 3 x 3 matrices. Find |— season. 6. [5 pts.] Let AX = B be a. linear system such that 2 —1 3 I 2 A: U 1 —2 .X= y .andB= U . —1 —1 1 2: 1 [3.) Find A—l, (b) Use part (a) to solve AX = B. 7. [5 13115.] Find all values of a for which the following system has: (i) no solution, (ii) unique solution. [iii) infinitely many solutions. III —— 2y — z = 2 23 + 5y + :5 = 5 I —— y -|— [o2 — 5}: = o 1 O. 09 Key Solutions First :M'ld-Term in LINEAR ALGEBRA (Ninth, 111} - 0 LLetA: I 5' .ThemAB: 1 y 1 “ = " w 2 —l. 3 u: 2. 2+2y QI—y _ 1 2 . ._ _ n_o [3+2w 23—36] — [2 4J.".[fhet'ta-fole,:IL —1:y—Ulv—-1and ._ . _ 1E} _. _1__L —l —2 w—U,1.e.A— 0 O:|.OI,SIIICEEIB — 5[_2 1],“ejust compute A = [.ABjB—L. 2. Since A and B are both :‘kew—:';\-'1'ri_met.rit:E AT 2 —A and ET 2 —B. Therefore: [_AB+BA}T = [ABIIT+ [521)?" = {BT){AT)+ LATHE?) = [—B)[_—A)+ [_—A][_—B) = BA + AB = AB + BA. 2 —2 —2 32 321-10 = I3 if and onlyiff." = 213-11 Thus, C = AB-1 = [ 10 —e. 12 J . 6 10 —13 4. a. If A is non—singulaL then 24—1" exists and 14—114 2 1'. Thus |A_]'| |A| = 1, which means that |A| can not be zero. 13. False: let A be any 21 X n nomingiular matrix. Then 3 = —A is also nonsingular. However, A +B = I] is not. Example: 11 = 2. A = I2. 5, |— 3A2[B2;IT| = [—3)3 |A2| “32m = —27 |A| |A| |B||B| = 42-4-4- —1 —1' T-—i—=—me. 67 For [_a]|, we have 243E100 100E121 [AII3:= o 1—25010 “WW 0105—2 —5 —4 —1—11E001 0015—1—3—2 [I3lA—l:: 3 and for [bl], we have X = A—lB. Thus: X = [ —8 ] . —-—i 1 2 —1 E 2 1 2 —1 E 2 7—[AIB]~ 2 5 1' 5 mm“ 0 1 35 1 11522—5521 OOaQ—IEa—l Therefore! we have no solution if a. = —1, unique solution if a. E RIDE—1! I}! and infinitely many solutions if a. = 1. l0 0. 09 ...
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111-2007&amp;amp;2008-2-M10-March2008_2 - Kuwait University...

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