111-1999&2000-3-F10-May2000

111-1999&2000-3-F10-May2000 - Kuwait University...

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Unformatted text preview: Kuwait University Department of Mathematics and Computer Science Math. 111, Linear Algebra Date: 25/5/2000 Final Exam- Time Allowed: 2 Hrs. M Answer all questions. Calculators and Mobile Phones are not allowed. 1. [14 pts‘] Answer each of the following as True or False (justify your answer): a. The angle between the two Vectors X 2 (1,0,0,0) and Y 2 (1,0. l, 1) is b. If X1, X2 and X3 are solutions of AX : B, then EEXI + ng -— X3 is a solution to AX = O. c. L: R3 -—v R2, given by L(a:,y, z) = (:5 + y.y2), is a. linear transformation. d. The set of all 2 x 2 singular matrices is a. subspace of szg (the space of all 2 x 2 matrices). e. If A is a 3 X 3 skew-symmetric matrix, then iAI = 0‘ - f. If A and B are n x n matrices such that A = BT, then I2A‘lBi 2 2. g. If A and B are similar n x n matrices, then |Al = 2. [5 pts.] Let X1 :: (1,0,—2), X2 = (l, —1,0) and X3 = (0,2,4). (a) Show that (1, 5, —4) is in spam {X1,X2,X3}. (b) Let L : R3 —» P2 be a linear transformation such that L(X1) = t2 + 3, L(X2) = t2 + z — 2, LfXa) :z+4. Find L(1,5, —4) 10—123 I 2 1 —2 2 1 3 [5 pts.iLetA»— 0 1 H5 8 U 1 -1 —1 4. 3 (3.} Find a basis for the null space of A. (b) Find the rank of A. 4. {6 pts.] Let L : P2 -+ P2 be a linear transformation defined by Lfot2 + bt + c} z (b - a)t2 + (a — cjt. (a) Is it? +3t — 2 in the range of L ? (b) Ist2+t+1 in the kernel OH)? (0) Find a basis for Ker L. (d) Is L one-to-one ? (a) (e) Is L onto?. Justify your answers in ail parts of this question. 2 3 0- 5.[6pts.]LetA= 10 0 5 5 ——1 (a) Find the eigenvalues of A. I (b) Find, ifpossible, a. nonsingular matrix P and a diagonal matrix D such that D = P“1AP. 6. {4 pts.] Let A and B be 71 x 7; matrices. Prove that (a) If A and B are nonsinguiar, then AB is nonsingular and (AB)‘1 = B'IA‘I. (b) If A is an upper triangular matrix, then the eigenvalues of A are the elements on the main diagonal of A. m\ 9 O -. Math 111, Linear Algebra 25 May, 2000, Final Exam. Solutions W Al. a. F, (0.5 pi“), c059 : fi implies 0 yé ’3—', (1.5 pt). b. F, (0.5 pt.).A(%X1+§Xi —X3) = B, (1.5 m). c. F, (0.5 1313.), LOG + X2) 96 L(Xi) + L(X2), (1.5 pt). 3 d. F, (0.5 pt.), Let W be the subset in the question. Then A : 0 0 L B :—= [i 2 0 I " 0 1 0 0 i u l W implies |A + BI = 2 v7]: 0. Hence A + B is not in W, (1.5 pt). 3. T, (0.5 m), A 2 —AT. (0.5 pt), implies (A; : | sATl implies ;A| = —-|A|, (0.5 m). Hence IA] : 0, (0.5 pt.). f. F, (0.5 pt.), I2A“1B| : 2“[A‘1|A| = 2“. (1.5 pt). g. T, (0.5 pt), B = P‘JAP, (0.5pt.) implies |A| = |BI, (1 pt). A2. (a) Let c1(1,0,—2)+02(1,fi1,0)+c;~,(0,2,4)=(1,5,—4), (0.5 pt.) 1 1 0 Cl 1 Form and solve the equivalent system 0 ‘—1 2 C2 = 5 , (2 pt.). *3 U 4 C3 —4 The solution of the system is c1 = 4, 02 = span, (0.5 pt.) (‘0) L(1,5, —4) : L(1,0, v—2) + L(1, -1,0) + L(0,2,4),(1pt.). L(i,5,—4) : 4(i2+3) u3(t2+t-—2)+t+4=t2~—2t+22, (1 p0). —3,ca : 1. Hence the given vector lies in the 1 O —l 2 3 1 0 0 0 2 0m 3 i .0. s .5 35?: Hit) 0 —1 0 2 5 0 0 0 0 0 The reduced system 021+ 2m5 = 0, :02 — 2:04 H 5035 = 0,033 — 234 ~ 035 = 0, set 55 = t and 0:4 :T,(1pt.). ' —2 0 5 2 Hence, X :t 1 +1“ 2 ,(1 pt.) 0 1 1 O —2 0 5 2 The set 1 , 2 is a basis for the null space of A, (1 pt). 0 1 1 0 (b) Rank of A = 5 — nullity of A m 5 i 2 2 3, (1pt.) (or any other- way). A4. (a) No, (0.5 pt), since t2 + 3t — 2 = (b i a)i2 + (a — c)t implies i2 = 0, (0.5 pt). (b) Yes, (0.5 pt), since L(t2+t+ 1) 2 O, (0.5 pt). (0) To find a. basis for kerL, we seek the solution of (b — a):2 + (a — c)t : O which is b: a,c = a, (1 pt). Hence kerL = {a(t2 + t+ 1) : a E R}. Hence, {t2 + t + l} is a basis for kerL, (ipt). (cl) No, (0.5pt), since liter}:- yé {0,3,} (0.5pt). (e) No, (0.5pt), since dim(rangeL) % 3 (0.5pt). OO 99 ...
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111-1999&2000-3-F10-May2000 - Kuwait University...

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