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111-2008&2009-2-F10-June2009

111-2008&2009-2-F10-June2009 - Kuwait University...

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Unformatted text preview: Kuwait University Department of Mathematics and Computer Science Math. 111, Linear Algebra Date: 18/6/2009 Final Exam. Time Allowed: 2 Hrs. Answer all questions. Calculators and Mobile Phones are not allowed. 1. [9 pts.[ Answer each of the following as True or False (justify your answer): a. If A and B are 2 x 2 matrices, then (AB)2 = AZBZ. b. If A is a nonsingular symmetric matrix with A = BT, then (ATBTV1 : (A‘1)2. The sign of the term a15a22034a41a53 in the expansion of a 5 x 5 determinant is +. d. The lineI T_ — —L-— ——_i-3- is parallel to the plane 4m + 10y —- 62— — 19. e. If X and Y are two orthogonal unit vectors in R”, then [IY -— X[[ = \/_. O f . The matrix A = [ [1) Bl ] is not diagonalizable. 1 +2 0 0 3 2 —5 —3 —2 6' 2. [6 pts.] Let A — 0 5 15 10 0 2 6 18 8 6 (21) Find a basis for the row space of A. (b) Find a basis for the column space of A that contains only columns of A. (c) What is the comm rank of A? 2c 6—3 3. [3 pts.] Find all values of c for which the matrix A = [ 1 c _ 3 J has zero nullity. 4. [3 pts.[ Find the equation of the plane passing through the point (2 ~3, 5) and contains the twolinesI—g—Z=%:l—_‘Tand$=2—t, yz~3+5t z=4+3t. 5. [3 pts.] Determine whether the set S = ((1, —1,2), (0,1, ~2),(1, l, —2), (2, —1,2)} spans R3. 6. [4 pts.[ Let S— — {X1,X2,X3} be a linearly independent set of R". Determine whether the set T: [X1 + X2 + X3, X1 + 2X2 + 3X3, X; + 2X3} is linearly independent or linearly dependent. 2 1 0 ~3 —2 4 «5 —3 -3 5 3 3 —4 6 7. [6 pts.] Let A = (a) Show that /\ z 1 is an eigenvalue of A. (b) Find the eigenvectors that are associated with the eigenvalue A = 1. 8. [6 p_ts.] Prove the following (a) If_A is an n x n nonsingular matrix and b is a positive real number, then adj(bA)_1 2 b1—” adj A“. (b) If A and B are two n x 71 similar matrices, then A and B have the same eigenvalues. CO Math 11], Linear Algebra, 18 June 2009, Final Exam., Solutions A1a.F, (0.5pt.,) Azh “11] 3:“ 1]. Hence, (AB)2: [g 2] #11232: [44 44] (1 pt.) A11). T, (05 pt) (ATBTr1 = (A2)-1 =(A‘1)2, (1pt.) Alc. F, (0.5 pt.), the number of inversions : 7, (1 pt.) A1d.F,(0.5pt),—62x4+10><5+( 6X) (— 3)7EO, (1pt.) Ale- T1(0'5 PL), llY - X“ = HXH + ”Y” = V5, (1th Alf. T, (0.5 pt), A 2 it, the eigenvalues-are'not real, (1 pt.) 1 0 0 —2 3 . 0 1 0 —1 0 A2a. the r.r.e.f. of A13 0 0 1 1 0 (2 pts). 0 0 0 0 0 A basis for the row space is {(1,0,0,— 2 ,,3) (0,1,0, —1,0), (0, 0 1 1 0)} (1.5 pt). I I ’ A2b. A basis for the column space is {(1,2,0,2), (~2, ~5,5,6), (0, —3,15.18)} (1.5 pt.) A2c. The column rank of A = 3 , (1 pt.) A3. Since, nullity of A : 0, then IA) 3A 0. Thus, 2c(c — 3) — (c — 3) 74 0 implies that c 75 3 or c gé 1/2. Hence, c E R/{3,1/2}., (3 pts.) 1 j k A4. The normal to the two given lines is 3 ~2 —1 2 —1 ~ 8j + 13k.. The equation —1 5 3 0f the plane is (:1: H 2) + 8(y + 3) —13(2 _ 5): 0., (3 pts.) A5. Let (a, b, c) E R3 and c1(1, —1,2) + (22(0, 1,2) + C3(1,1,2) +C4(2, ~1,2=) (a, b, c). This is equivalent to a non— hom. sys. and we solve it as follows 1012Ea 1012Ea —11 1—151; ”a 0121Eb+a 2—2—2251: 000052b+c This system has a. soln. if 217 + c = 0. Thus the set 8 doesn’t span R3. A6. Let . C] (X1 + X2 + X3) + 62(X1+ 2X2 + 3X3) + C3(X2 ‘1‘ 2X3) 2 0. Hence, (61 + C2)X] + (C1 + 2C2 + C3)X2 + (Cl + 3C2 + 2C3)X3 I 0. Since, 1 O :0 COMB 1 2 Wm Hence, the set T is linearly dependent. m1~10 *1 A721. l1 — A| : g 3 :2 g = 0, which means that A = 1 is an eigenvalue of A. ~73 —3 4 —5 1 l 0 1 . O O 2 -—1 _ I A7b. The matrlx I v A w 0 O 0 {J . Solvmg the system (I k A)X :_O, 0 0 O O we obtain the eigenvectors X1 = (—1,1,0,0)T and X2 = (—2,0,1,2)T. ABa. adj(bA)‘1 = \(bA]*1]([bA)‘1)—1 : bf"|A’1}b(A’1}’1=b“"adj(A‘1). ASb. Proved in Class. O. CO ...
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