Introduction to Linear Algebra练习题答案

1、INTRODUCTIONTOLINEARALGEBRAThird EditionMANUAL FOR INSTRUCTORSGilbert SMassachusetts Institute of Technology/18.06/www/gsWellesley-Cambridge PressBox 812060Wellesley, Massachusetts 02482Solutions to ExercisesProble

2、m Set 1.1, page 61 Line through (1, 1, 1); plane; same plane!3 v = (2, 2) and w = (1, 1).4 3v + w = (7, 5) and v 3w = (1, 5) and cv + dw = (2c + d, c + 2d).5 u + v = (2, 3, 1) and u + v + w = (0, 0, 0) and 2u + 2v + w = (add first answers) = (2, 3, 1).6 The components of every cv + dw add to zero. C

3、hoose c = 4 and d = 10 to get (4, 2, 6).8 The other diagonal is v w (or else w v). Adding diagonals gives 2v (or 2w).9 The fourth corner can be (4, 4) or (4, 0) or (2, 2).10 i + j is the diagonal of the base.11 Five more corners (0, 0, 1), (1, 1, 0), (1, 0, 1), (0, 1, 1), (1, 1, 1). The center point

4、 is ( 12 , 12 , 12 ). The centers of the six faces are ( 12 , 12 , 0), ( 12 , 12 , 1) and (0, 12 , 12 ), (1, 12 , 12 ) and ( 12 , 0, 12 ), ( 12 , 1, 12 ).12 A four-dimensional cube has 24 = 16 corners and 2 4 = 8 three-dimensional sides and 24 two-dimensional faces and 32 one-dimensional edges. See

5、Worked Example 2.4 A.13 sum = zero vector; sum = 4:00 vector; 1:00 is 60 from horizontal = (cos 3 , sin 3 ) = ( 12 , 23 ).14 Sum = 12j since j = (0, 1) is added to every vector.15 The point 34 v + 14 w is three-fourths of the way to v starting from w. The vector 14 v + 14 w ishalfway to u = 12 v + 1

6、2 w, and the vector v + w is 2u (the far corner of the parallelogram).16 All combinations with c + d = 1 are on the line through v and w. The point V = v + 2w is on that line beyond w.17 The vectors cv + cw fill out the line passing through (0, 0) and u = 12 v + 12 w. It continues beyond v + w and (

7、0, 0). With c 0, half this line is removed and the “ray” starts at (0, 0).18 The combinations with 0 c 1 and 0 d 1 fill the parallelogram with sides v and w.19 With c 0 and d 0 we get the “cone” or “wedge” between v and w.20 (a)1u +1v +1w is the center of the triangle between u, v and w;1u +1w is th

8、e center32332of the edge between u and w(b) To fill in the triangle keep c 0, d 0, e 0, andc + d + e = 1.3421 The sum is (v u) + (w v) + (u w) = zero vector.22 The vector 12 (u + v + w) is outside the pyramid because c + d + e = 12 + 12 + 12 1.23 All vectors are combinations of u, v, and w.24 Vector

9、s cv are in both planes.25(a) Choose u = v = w = any nonzero vector and w to be a combination like u + v.(b) Choose u and v in dierent directions,26 The solution is c = 2 and d = 4. Then 2(1, 2) + 4(3, 1) = (14, 8).27 The combinations of (1, 0, 0) and (0, 1, 0) fill the xy plane in xyz space.28An ex

10、ample is (a, b) = (3, 6) and (c, d) = (1, 2). The ratios a/c and b/d are equal. Then ad = bc. Then (divide by bd) the ratios a/b and c/d are equal!Problem Set 1.2, page 171 u v = 1.4, u w = 0, v w = 24 = w v.2 kuk = 1 and kvk = 5 = kwk. Then 1.4 (1)(5) and 24 (5)(5).3 Unit vectors v/kvk = ( 35 , 45

11、) = (.6, .8) and w/kwk = ( 45 , 35 ) = (.8, .6). The cosine of isvw= 24 . The vectors w, u, w make 0, 90, 180 angles with w.kvkkwk254 u1 = v/kvk = 110 (3, 1) and u2 = w/kwk = 13 (2, 1, 2). U 1 = 110 (1, 3) or 110 (1, 3). U 2could be 1 (1, 2, 0).55(a) v (v) = 1(b) (v +w) (v w) = v v +w v v w w w = 1+

12、( )( )1 = 0so = 90(c) (v 2w) (v + 2w) = v v 4w w = 36(a) cos =1so = 60 orradians(b) cos = 0 so = 90 orradians(2)(1)32(2)(2)1+31so = 60 or(d) cos = 1/so = 135 or3.(c) cos =22347 All vectors w = (c, 2c); all vectors (x, y, z) with x + y + z = 0 lie on a plane; all vectors perpendicular to (1, 1, 1) an

13、d (1, 2, 3) lie on a line.8 (a) False(b) True: u (cv + dw) = cu v + du w = 0(c) True9 If v2w2/v1w1 = 1 then v2w2 = v1w1 or v1w1 + v2w2 = 0.10 Slopes 21 and 12 multiply to give 1: perpendicular.11 v w 90; this is half of the plane.12 (1, 1) perpendicular to (1, 5) c(1, 1) if 6 2c = 0 or c = 3; v (w c

14、v) = 0 if c = v w/v v.13 v = (1, 0, 1), w = (0, 1, 0).14 u = (1, 1, 0, 0), v = (0, 0, 1, 1), w = (1, 1, 1, 1).151(x + y) = 5; cos = 216/1010 = .8.216 kvk2 = 9 so kvk = 3; u = 13 v; w = (1, 1, 0, . . . , 0).22222222, cos + cos + cos= 1.17 cos = 1/2, cos = 0, cos = 1/ = (v1+ v2+ v3 )/kvk518 kvk2 = 42

15、+ 22 = 20, kwk2 = (1)2 + 22 = 5, k(3, 4)k2 = 25 = 20 + 5.19 v w = (5, 0) also has (length)2 = 25. Choose v = (1, 1) and w = (0, 1) which are not perpendicular; (length of v)2 + (length of w)2 = 12 + 12 + 12 but (length of v w)2 = 1.20 (v +w)(v +w) = (v +w)v +(v +w)w = v (v +w)+w (v +w) = v v +v w +w

16、 v +w w = v v + 2v w + w w. Notice v w = w v!21 2v w 2kvkkwk leads to kv + wk2 = v v + 2v w + w w kvk2 + 2kvkkwk + kwk2 = (kvk + kwk)2.22 Compare v v + w w with (v w) (v w) to find that 2v w = 0. Divide by 2.23 cos = w1/kwk and sin = w2/kwk. Then cos(a) = cos cos +sin sin = v1w1/kvkkwk+ v2w2/kvkkwk

17、= v w/kvkkwk.24 We know that (v w) (v w) = v v 2v w +w w. The Law of Cosines writes kvkkwk cos for v w. When 90 with each other: (1, 0), (1, 4), (1, 4). Four vectors could not do this (360 total angle). How many can do this in R3 or Rn?Problem Set 1.31 (x, y, z) = (2, 0, 0) and (0, 6, 0); n = (3, 1,

18、 1); dot product (3, 1, 1) (2, 6, 0) = 0.24x y 2z = 1 is parallel to every plane 4x y 2z = d and perpendicular to n = (4, 1, 2).3(a) True (assuming n 6= 0)(b) False(c) True.4(a) x + 5y + 2z = 14(b) x + 5y + 2z = 30(c) y = 0.5 The plane changes to the symmetric plane on the other side of the origin.6

19、 x y z = 0.7 x + 4y = 0; x + 4y = 14.8 u = (2, 0, 0), v = (0, 2, 0), w = (0, 0, 2). Need c + d + e = 1.69 x + 4y + z + 2t = 8.10 x 4y + 2z = 0.11 We choose v0 = (6, 0, 0) and then in-plane vectors (3, 1, 0) and (1, 0, 1). The points on the plane are v0 + y(3, 1, 0) + z(1, 0, 1).12 v0 = (0, 0, 0); al

20、l vectors in the plane are combinations y(2, 1, 0) + z( 12 , 0, 1).13 v0 = (0, 0, 0); all solutions are combinations y(1, 1, 0) + z(1, 0, 1).14 Particular point (9, 0); solution (3, 1); points are (9, 0) + y(3, 1) = (3y + 9, y).15 v0 = (24, 0, 0, 0); solutions (2, 1, 0, 0) and (3, 0, 1, 0) and (4, 0

21、, 0, 1). Combine to get (24 2y 3z 4t, y, z, t).16 Choose v0 = (0, 6, 0) with two zero components. Then set components to 1 to choose (1, 0, 0)and (0, 3/2, 1). Combinations are (x, 6 32 z, z).17 Now |d|/knk = 12/ 56 = 12/2 14 = 6/ 14. Same answer because same plane.18(a) |d|/knk = 18/3 = 6 and v = (4

22、, 4, 2)(b) |d|/knk = 0 and v = 0(c) |d|/knk = 6/2 and v = 3n = (3, 0, 3).19(a) Shortest distance is along perpendicular to line(b) Need t + 4t = 25 or t = 5(c) The distance to (5, 10) is 125.20(a) n = (a, b)(b) t = c/(a2 + b2)(c) This distance to tn = (ca, cb)/(a2 + b2) is|c|/.a2 + b221 Substitute x

23、 = 1 + t, y = 2t, z = 5 2t to find (1 + t) + 2(2t) 2(5 2t) = 27 or 9 + 9t = 27 or t = 4. Then ktnk = 12.22Shortest distance in the direction of n; w + tn lies on the plane when n w + tn n = d or t = (d n w)/n n. The distance is |d n w|/knk (which is |d|/knk when w = 0).23 The vectors (1, 2, 3) and (

24、1, 1, 1) are perpendicular to the line. Set x = 0 to find y = 16 and z = 14. Set y = 0 to find x = 9/2 and z = 5/2. These particular points are (0, 16, 14) and (9/2, 0, 5/2).24 (a) n = (1, 1, 1, 1)(b) |d|/knk = 12(c) dn/n n = ( 14 , 14 , 14 , 14 )(d) v0 = (1, 0, 0, 0) (e) (1, 1, 0, 0), (1, 0, 1, 0),

25、 (1, 0, 0, 1)(f) all points (1 y z + t, y, z, t).25 n = (1, 1, 1) or any nonzero (c, c, c). 26 cos = (0, 1, 1) (1, 0, 1)/2 2 = 12 so = 60.Problem Set 2.1, page 291 The planes x = 2 and y = 3 and z = 4 are perpendicular to the x, y, z axes.2 The vectors are i = (1, 0, 0) and j = (0, 1, 0) and k = (0,

26、 0, 1) and b = (2, 3, 4) = 2i + 3j + 4k.73 The planes are the same: 2y = 6 is y = 3, and 3z = 12 is z = 4. The solution is the same intersection point. The columns are changed; but same combination xb = x.4 The solution is not changed; the second plane and row 2 of the matrix and all columns of the

27、matrix are changed.5 If z = 2 then x + y = 0 and x y = z give the point (1, 1, 2). If z = 0 then x + y = 6 and x y = 4 give the point (5, 1, 0). Halfway between is (3, 0, 1).6 If x, y, z satisfy the first two equations they also satisfy the third equation. The line L ofsolutions contains v = (1, 1,

28、0) and w = ( 12 , 1, 12 ) and u = 12 v + 12 w and all combinations cv + dw with c + d = 1.7 Equation 1 + equation 2 equation 3 is now 0 = 4. Solution impossible.8 Column 3 = Column 1; solutions (x, y, z) = (1, 1, 0) or (0, 1, 1) and you can add any multiple of (1, 0, 1); b = (4, 6, c) needs c = 10 f

29、or solvability.9 Four planes in 4-dimensional space normally meet at a point. The solution to Ax = (3, 3, 3, 2) is x = (0, 0, 1, 2) if A has columns (1, 0, 0, 0), (1, 1, 0, 0), (1, 1, 1, 0), (1, 1, 1, 1).10 Ax = (18, 5, 0), Ax = (3, 4, 5, 5).11 Nine multiplications for Ax = (18, 5, 0).12 (14, 22) an

30、d (0, 0) and (9, 7).13 (z, y, x) and (0, 0, 0) and (3, 3, 6).14(a) x has n components, Ax has m components (b) Planes in n-dimensional space, but the columns are in m-dimensional space.15 2x + 3y + z + 5t = 8 is Ax = b with the 1 by 4 matrix A = 2 3 1 5 . The solutions x filla 3D “plane” in 4 dimens

31、ions.16 I =1 0 , P = 0 1.011017 R =01, 180rotation from R2 = 10=I.1 00 101 000118 P =001produces (y, z, x) and Q =100recovers (x, y, z).10001019 E = 1 0, E =1001 1 0 .1100120 E =1 0 0, E1=1 0 01Ev = (3, 4, 5).010010, Ev = (3, 4, 8), E10110121 P1 = 1 0 , P2 =0 0 , P1v =5 , P2P1v =0 .0 00 10082222R =1

32、.222x23The dot product 14 5 y= (1 by 3)(3 by 1) is zero for points (x, y, z) on a plane inzthree dimensions. The columns of A are one-dimensional vectors.24A = 1 2;3 4 and x = 52 0 and b = 17 0. r = b A x prints as zero.25A v = 345 0 and v0 v = 50; v A gives an error message.26ones(4, 4) ones(4, 1)

33、= 4 4 44 0; B w = 1010 10 10 0.27 The row picture has two lines meeting at (4, 2). The column picture has 4(1, 1) + 2(2, 1) = 4(column 1) + 2(column 2) = right side (0, 6).28 The row picture shows 2 planes in 3-dimensional space. The column picture is in 2-dimensional space. The solutions normally l

34、ie on a line.29 The row picture shows four lines. The column picture is in four-dimensional space. No solutionunless the right side is a combination of the two columns.30 u2=.7, u3=.65. The components always add to 1. They are always positive.3.3531 u7, v7, w7 are all close to (.6, .4). Their compon

35、ents still add to .6=.6= steady state s. No change when multiplied by. .8 3 45 + u5 u + v5 v34M =159=5 u v55 + u + v; M3(1, 1, 1) = (15, 15, 15);6725 + v5 + uv5uM4(1,1, 1, 1) = (34, 34, 34, 34) because the numbers 1 to 16 add to 136 which is 4(34).Problem Set 2.2, page 401 Mult

36、iply by l = 102 = 5 and subtract to find 2x + 3y = 14 and 6y = 6.2 y = 1 and then x = 2. Multiplying the right side by 4 will multiply (x, y) by 4 to give the solution (x, y) = (8, 4).3 Subtract 12 times equation 1 (or add 12 times equation 1). The new second equation is 3y = 3.Then y = 1 and x = 5.

37、 If the right side changes sign, so does the solution: (x, y) = (5, 1).4 Subtract l = ac times equation 1. The new second pivot multiplying y is d(cb/a) or (adbc)/a. Then y = (ag cf)/(ad bc).5 6x + 4y is 2 times 3x + 2y. There is no solution unless the right side is 2 10 = 20. Then all points on the

38、 line 3x + 2y = 10 are solutions, including (0, 5) and (4, 1).96 Singular system if b = 4, because 4x + 8y is 2 times 2x + 4y. Then g = 2 16 = 32 makes the system solvable. The lines become the same: infinitely many solutions like (8, 0) and (0, 4).7 If a = 2 elimination must fail. The equations hav

39、e no solution. If a = 0 elimination stops for a row exchange. Then 3y = 3 gives y = 1 and 4x + 6y = 6 gives x = 3.8 If k = 3 elimination must fail: no solution. If k = 3, elimination gives 0 = 0 in equation 2:infinitely many solutions. If k = 0 a row exchange is needed: one solution.9 6x 4y is 2 tim

40、es (3x 2y). Therefore we need b2 = 2b1. Then there will be infinitely many solutions.10 The equation y = 1 comes from elimination. Then x = 4 and 5x 4y = c = 16.11 2x + 3y +z = 8x = 2y +3z = 4givesy = 18z = 8z = 1If a zero is at the start of row 2 or 3,that avoids a row operation.12 2x 3y = 32x 3y =

41、 3x = 3Subtract 2 row 1 from row 2y +z = 1givesy +z = 1and y = 1Subtract 1 row 1 from row 32y 3z = 25z = 0z = 0Subtract 2 row 2 from row 313 Subtract 2 times row 1 from row 2 to reach (d 10)y z = 2. Equation (3) is y z = 3. If d = 10 exchange rows 2 and 3. If d = 11 the system is singular; third piv

42、ot is missing.14 The second pivot position will contain 2 b. If b = 2 we exchange with row 3. If b = 1 (singular case) the second equation is y z = 0. A solution is (1, 1, 1).150x + 0y + 2z = 40x + 3y + 4z = 4(a)x + 2y + 2z = 5(b)x + 2y + 2z = 50x + 3y + 4z = 60x + 3y + 4z = 6(exchange 1 and 2, then

43、 2 and 3)(rows 1 and 3 are not consistent)16 If row 1 = row 2, then row 2 is zero after the first step; exchange the zero row with row 3 and there is no third pivot. If column 1 = column 2 there is no second pivot.17 x + 2y + 3z = 0, 4x + 8y + 12z = 0, 5x + 10y + 15z = 0 has infinitely many solution

44、s.18 Row 2 becomes 3y 4z = 5, then row 3 becomes (q + 4)z = t 5. If q = 4 the system is singular no third pivot. Then if t = 5 the third equation is 0 = 0. Choosing z = 1 the equation 3y 4z = 5 gives y = 3 and equation 1 gives x = 9.19 (a) Another solution is1(x + X, y + Y, z + Z).(b) If 25 planes meet at two points, they2meet along the whole line through those two points.20 Singular if row 3 is a combination of rows 1