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111-2006&amp;2007-3-M20-July2007

# 111-2006&amp;2007-3-M20-July2007 - Kuwait University...

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Unformatted text preview: Kuwait University, Faculty o—i‘_Science Dept. of Mathematics & Cornputer Science LINEAR ALGEBRA (Math 111) Second Mid-Term Test July 17, 2007 Duration 90 minutes Calculators and mobile phones are NOT allowed Answer the following questions: 1 0 0 l O —1 1 —l 1. [2+2 pts.] Let A — 1 0 1 3 ‘2. 4 —4 7 (a) Find lAl. (b) Use part (a) to show that 1::de is nonsingular and ﬁnd (ode)"1. 0.1 012- I 33 C1 + 451 2131 — 3&1 20.} +51 2. [3 ptS.] Let C = b1 52 b3 and D 3 Cg +452 2132 ~— 3112 2&2 +52 . (31 Cg 63 C3 + 453 253 — 3&3 2(13 4- 53 101: 2, ﬁnd lDl. 3. [3 pts.] Find a vector Y E R4 which is orthogonal to each of the vectors X1 = (1,1,1,1),X2 = (1,1,2,3), X3 = (1,1,2,4) and has the length HYH : 5\/§. 4. [-l+2+2+'2 pts.] Let the line L : :1: :1-l—2t, y = 4-15, 3 2 5—1:, the points P(0,0, 1) and Q(l, 1,2), and the plane H : 4m — 2y — 22 +1: 0. (a) Show that the line L is perpendicular to the plane H. (b) Find parametric equations of the line L’ passing through the points P and Q. (0) Determine whether the lines L and L’ intersect. Write an equation for the plane which contains the line L and the point P. 5. [2 pts.] let V be the set of ordered pairs of real numbers (33,30 with operations: (x. y) e (33’, y’) = (I + w’. :9 +10 and 60 (may) = (cm. ytc E R- Is (V. a G) a vector space? Justify your answer. ' 6. [6 pts.] Answer each of the following as True or False. Justify your answer. (a) If A is a '3 >63 matrix and (1de = 0, then A = O. - (b) If A and B are 5 X 5 matrices and AB = —BA, then Mi 2 0 or lBI = O. _ If H3X + 2V“ 2 “2X + 33’“, then = l d) The intersection of the planes ac —— 23; + z + l = 0 and 2:1: — 4y + 2;: + l = 0 is a line.' “WW .0 July 17, 2007 Math 111 Solution WW 0 0 O l 1 I --1 1 (sum: 0 ‘1 1‘1 = 0 1 2 2—1. 1 0 1 2 4 _4 5 2 4 —4 5 (b) |adel:|AJ3=—1#0==>(ade)‘1=FhA=—A. Cl —7G1 2a1+b1 Cl £11 51 (21 0.1 151 2. 624-452 —7a2 2a2+bg =—7 C2+452 G2 bg "—2—? C2 (12 52 = Cg'l‘élbg —7a3 2613-1-53 (33-1-4513 a3 I53 03 a3 be, 410] = —14. 3. Y satisfy the equations: m+y+z+w = 0,0:+y+22+3w O 23+y+22+4w : O, with r111110 110010 solution: 1 1 2 3 | 0 ~ mm 0 0 1 0 1 0 =>Y=(—t,t,0,0). l112410l l0001|0 (ll/H = 5\/§ :> 2:2 = 50 => '1 = 1:5. Thus, Y1 = (—5,510,0),Y2 = (5, —5,0,0) are solutions. 4. (a) U :(2,—~1,—1)for L and the normal to the pane N = (4, —2, —2), N = 2U. (b) L’:x=r, yr'r, z:1+r. (c) LﬂL’ :1": 1+2t,r :4—t,1+r= 5—t:> 7": 3,t = 1, uniquesolution oz}; tWO lines intersect at the point (3, 3, 4). (d) The 'plane is determined by the two lines L and L’, so a normal Vector to the plane is (2, —1, —1) X (1, 1,1) : (0,—3,3) 22> —3(y — 3) + 3(2 — 4) = 0 => y~z+1=0 5. It is not because: (c+d)®(m,y) 2 (cm+d2:,y) 75 (cx+d:c, 2y) = c®(\$, y)®d®(sc,y), y i 0. . . 1 1 1 6. (a) False, A = 1 1 1 . l 1 1 (b) '1‘rue,-lAB(~=]—- BAI 2» |AB| = (*1)5|BA( => 2|AB| = 0 => M] = 0 or 131: e. (c, True, [BX + 2Y||2 = ||2X + 31/“2 => 9||X|12 + 12X - Y + 4i|YHE : 4”th2 + 12X - Y+ 91|Y|l2 => 1le1 = llYH- ' ((1) False, two planes 3: — 2y + z + 1 = 0 and 2&2 — 43,; + 22 + 1 = O are parallel. OO 99 ...
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111-2006&amp;2007-3-M20-July2007 - Kuwait University...

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