111-2007&2008-3-M20-July2008

111-2007&2008-3-M20-July2008 - Kuwait University...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Kuwait University Department of Mathematics and Computer Science Math. 111, Linear Algebra 15 July 2008 Second Exam. Total marks: 25 Duration: 90 min. m Answer all questions. Calculators and Mobile Phones are not allowed. 1. (4pt) (a) Suppose A is a 4 x 4 matrix with |A| = —2. Find |adj(A)[ and ladj(adj(A))[. l —2 0 0 (b) For the matrix A z i; _2 3 8 find the cofactor A23. 0 0 O l 2. (4pt) Let P = (1,2,1),Q = (—1,1,3) and R : (—2, —2, *2). Find an equation of i the plane through the points 13,62 and R. 3. (3pt) Find parametric equations of the line through the points (1,—2.4) and [6, 2, *3). Determine Whether the point (1. 1, 1) is on this line. 4. (3pt) Suppose that X is a vector in R3 such that X x U = O for every-unit vector U. Prove that X = O. 5. (Spt) Suppose U and V are non—zero vectors and HU X V|| : U - V. Find the angle between U and V. 6. (4pt) Prove that if U and V are vectors in R”, then “U + V” g + IIVIF. 7. (4pt) Determine whether the following subsets are subspaces of R3 (a) W={(a,b,c) 6R3za+b=2c}. (b) W:{(a,b,c)ER3:ac=0}. OO 99 Math. 111, Linear Algebra, Second Exam solutions, 15 July 2008 l. (4pt) (a) Iadj(A)l : |A|3 : is and |adj(adj(A))| = |adj(A)|3 : -83, 1 —2 0 (b) «423:7 3 77 0 =—(~1)=1_ 0 0 1 2. (4pt) Put u = PQ = (—2, $1,211; : PR = (73, M4, ~3). Then a normal vector to the plane is o X t; z (11, —12,5), so that an equation is 11(z—1)712(y~2)+5(z—1)=0, or 113:—12y+52+8=0. (Or: the plane has equation ax + by + C2 + d = O and since P lies on the plane, a + 213+ c + d = 0. Similarly for Q, R. Solve the resulting system for a, b, c.) 3. (Bpt) u z (6, 2,—3) — {1-,-—2,-4)- = (5,4, —7) is in the direction of the line, so the equations are m : 1+ 5mg = —2 + 4t,z = 4 — 7t. . If the point were on the line, we would have 1 = 1 + 51:, so i : 0. But then 3/ : —-2 7E 1, so the point is not on the line. 4. (3pt) Put X z (a,b,c). Then Xxi:(0,c,—b)=0,sothetb:c:0a.nd X xj= (70,0,a)=0, so that 01:0. ThusXIO. (Or: leXi:0,thenX=ti. IfXXj=0,thenX=t‘j andtirt'j,sz =0.) 5. (3pt) we have sine = ||U||Hlf|loost§l with HUII, 3% 0. Thus sinB 2 cost? and 6 2 17/4. 6. (4pt) Hu+ng= (n+o)-(u+v) =u-u+uvv+v-u+v:v :llull2+21¢‘v+llvll2 5 (using Cauchy—Schwartz) Hull2 + 2|luil|lv|l + Hull2 2 + Taking the positive square root gives the result. 7- (4m) (31) Take it = (a, b, c),u’ : (of, b’,c’) E W and r 6 1R. Then o+b = 2c,o’+b’ = 20’. Thus u+u’ = (a+o’,b+b’,c+c’) E W, since a+a’+b+b’ = 2(c+c’). Also, TU. = (TG,Tb, TC) 6 W, since m + rb = 2m. Thus W is a. subspace. (b) u : (1,0,0),n = (0,0,1) 6 l/V, but 11+?) : (1,0,1) 6 W, so not a subspace. ...
View Full Document

Page1 / 2

111-2007&2008-3-M20-July2008 - Kuwait University...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online