MATH 218 Problems Solution Manual Chapter 2.2, Fall 2011

46 Chapter 2 Linear Combinations and Linear Independence Uy V4 ‘| Ue v2 36. Let u= . v= . , and c a scalar, Then Un Un Uy + V1 cu, + Cv, Cul cUy U2 + U2 Cug + Cvg cug | v9 clu+v) =e . = = + =cu+ev, Un + Un CUn + CUn, CUn CUn, Uy Ug 37, Letu= . and ¢ and d scalars, Then Un, (c+ d)ur “PT cur + duy cut duy (e+ d)us Cug + duly Cu dug (c+d)ju= =| . = J fet. =cu+ du. (c+ d)un Clin + dun, CUn L dun [ws | duy cduy U2 dug cdug 38. Letu=| , and cand d scalars. Then c(du)=c}] , | = . = (ced)u. L Un | dun Cdn, [ts -] (1)tu ue » (L)ug 39. Let u= . . Then (b)u = , =u. Lun | (1)tn, 40. Suppose u+ 21 = u and u+22 =u. Then u+ 21 = u-+ Ze and hence, 21 = 29. Exercise Set 2.2 The vectors in IR? and R', called the coordinate vectors, are the unit vectors that define the standard axes, These vectors are 1 0 1 0 0 a=[5]e-[1] in R? andey=|0],e.=]| 1 |,eg= | 0 | in RS 0 0 . 1 with the obvious extension to R”. Every vector in the Euclidean spaces is a combination of the coordinate vectors. For example, V1 1 0 0 vg | =u] O | +ue] 1 | +23 | 0 U3 0 0 1 The coordinate vectors are special vectors but not in the sense of generating all the vectors in the space, Many sets of vectors can generate all other vectors. The vector v is a linear combination of V1,Va,.0+,Vn if there are scalars ¢1,¢2,...,, such that V = CyVy + CQVQ + + CnVne i i 1 i 2.2 Linear Combinations 47 For specific vectors we have already seen that an equation of this form generates a linear system. If the linear system has at least one solution, then v is’a linear combination of the others. Notice that the set of all linear combinations of the vectors e, and eg in R® is the ay plane and hence, ‘not all of IR. Similarly, all linear combinations of one nonzero vector in R? is a line, In the exercises when asked to determine whether or not a vector is a linear combination of other vectors, first set up the equation ‘above and then solve the resulting 1 0 linear system. For example, to determine all vectors in R® that are a linear combination of | 1 },| 1 ; 1 -1 2 a and 5 let v= | b |, be an arbitrary vector. Then form the augmented matrix —1 c ‘10 24a 1 0 2} a 1 oi 5 | 6 | that row reducestg]| 0 1 3 —a+b 1 ~1 -lle 0 0 0] -2a+b-e Hence, the components of any vector v that is a linear combination of the threé vectors must satisfy ~2a + b-c=0. This is not all the vectors in R°. In this case notice that 2 1 0 5 | =2]/1)4+3] 1 -1 _ -1 M@ Solutions to Exercises 1, To determine whether or not a vector is a linear combination of other vectors we always set up a vector equation of the form v = ¢ + v1 ++:+CnVn and then determine if the resulting linear system can be solved. In matrix form the linear system is 1 -2|-4 1 0/2 . a. . aus _ _ , 1 3 | ul that row reduces tg 01/3 , which has the unique solution cy = 2 and cg = 3. Hence, the vector v is a linear combination of vz and ve. . . —1 3) 18 1 0} -1 F . 2, Since the resulting linear system 2-9} —2 | to reduces ta oila)> which has the unique solution ¢; = ~1, cz = 4, then the vector v is a linear combination of vz and v2. 3. Since the resulting linear system ? 3 i row reduces t 0 “Be 2 | , it is inconsistent, then . v is not a linear combination of vz and ve. os 4, Since the resulting linear system 9 wa | ; row reduces to 4 a | , it is inconsistent, then v is not a linear combination of vz and ve. 5, Since the resulting linear system —-2 1|~-8 1 0;2 3 4] 10 | row reduces tg} 0 1] 1 4 2] 10 0 0/0 it is consistent with the unique solution c; = 2 and ce = 1, then v is a linear combination of v1 and vo. 6. Since the resulting linear system 38 2 ].-2 1 0} ~2 4 7] 6 | rowreducesto| 0 1] 2 —-1 3] 8 0 0} 0 48 , Chapter 2 Linear Combinations and Linear Independence it is consistent, then v is a linear combination of v1 and vg. 7, Since the resulting linear system 2 38 -2)2 1:0 0} -4 -2 0 O18 |—]0 1 0/2/3], 0 -38 +1) 2 0 0 1] -4 is consistent and has a unique solution, then v can be written in only one way as a linear combination of V1, V2, and v3. 9. Since 1 -1 0} -1 1 0 0/0 2 -1 1} il —};0 1 1/0], -l 38 2) 5 0 0 0jf1 the linear system is inconsistent, so v can not be written as a linear combination of v1, v2, and vg. 11. Since 2 1 -1) 3 -38 6 -1|-17 4 -1 2] 17 1 2 8 7 row reduces to 10 0} 3 0 ft 0}; -1 0 0 1/2 |? 0 0 0} 0 the linear system has a unique solution, so v can be written as a linear combination of v1, v2, and V3. 0 -1/3 1 0 ; | 0 reduces tg 0 1 4: 3 13. Since 1-1 2 Cl = 1+ fos, ¢2 = 14+ 403,09 ER. -2 3 -1 0 l 14. Since A - | sstuos to 0 scalars can be selected so v is a linear combination of v V1,,V2, and vg. Specifically, any set of scalars given by C1 = = 03, Cg = 3 + 3,03 ER. 0 -2 -2 2] 0 15. Since | 1 -1 -—3 ~—1]} —1 | reduces t 1 2 -1 -2/]-3 —1/3 7/3 can be selected so v is a linear combination of v4, v: 0 8. Since 1 -2 385 1 -1 -1 -1)/-4 | — | 0 0-1 -3|=7 0 oO HOO nee is consistent and has a unique solution, then v can be written in only one way as a linear combination of v1, V2, and vg. 10. Since -3 1 -1/-3 1 0 1/0 2 4 10) 5 —|0 1 2/0], 1 1 3 5 0 0 Oj}! the linear system is inconsistent, so v can not be written ag a linear combination of vz, v2, and vg. 12, Since 2 1 3/16 1 0 0/0 8 -l 1)3 fe. 0 1 0/0 4 2 -313 reduces t 00 1lo}? 5 8 1/7 0 0 0f1 the linear system is inconsistent, so v can not be written as a linear combination of vi, v2, and vg. 1} 1/3 1| 2/3 10 0 -6] 3 1 0 1 | -2 |; there are infinitely many ways in 00 1 ~-2) 2 , there are infinitely many ways in which scalars 2, and vg. Specifically, any set of scalars given by iF there .are infinitely many ways in which which scalars can be selected so v is a linear combination of v1, v2, Vs, a and v4. Specifically, any set of scalars given by cy = 3 + 6c4, cg = ~2 ~ c4,¢3 = 24 2cy,c4 ER. -1 0 O -3| 3 100 8 37 16. Since | —1 .—1 ~—1 ~—1]—-3 | reducesto| 0 1 0 -—12]—5 |, there are infinitely many ways 2 -1 -2 -2] 1 00 1 10] 5 in which scalars can be selected so v is a linear combination of v1, V2, Vs, and v4. Specifically, any set of scalars given by cy == 3 — 8¢4,¢c2 = —5 + 12c4,c3 = 5 — 10cq4,c4 € R. 2.2 Linear Combinations 49 17, The matrix equation 1 2 —2 3 -1 3 cis + caMa + aM = es | i]+a| l i] tel 9 ‘]=| 40 leads to the augmented matrix 1-2 -1]-2 10 2 3 8/4 a | 0 1 .0f-1 1 1 9 4 , Which reduces to 00113 |? -1 4 1/0 0 0 0) 0 and hence, the linear system has the unique solution c; = —1,cp = —1, and cg = 3. Consequently, the matrix M is:a linear combination of the three matrices. 18. Since the augmented matrix 2-1 1/2 10 0} 2 2 1 213 reduces to O01 0) 1 162 87/1 |O—-->/0 0 Lf-1]? 1 1 142 0 0 0} 0 the linear system has a unique solution, Consequently, the matrix M is a linear combination of the three matrices. 19, The matrix M is not a linear combination of the three matrices since 2 3 38 | 2 1 0 0}0 2-1 -1) 1 | aces to | 2 F 9] 0 -1 2 2 ]—1 | PEM Oo) 9 0 ilo}? 3-2 2 4} 2 00 0);1 so that the linear system is inconsistent. 1 0 0)2 10 0/0 ‘ . . a) matrt 0 1 071 ae 0 1 0}0 ta lt cpakamy tat 20. Since the augmented matrix 0 0 013 reduces tg 00110 |? the linear system is incon- -1 0 1/4 00 O}1 sistent. This is also immediately evident from row three of the original can not be written as a linear combination of M), M2, and M3. atrix. Consequently, the matrix M 5 1 3 1 2 —1 a. ax=2/ 1] =[ 9] 22, Ax=—| 2 |~|3]+43] 4 3B 2 1 23, The matrix product is AB = - i ; ; = u ie | , and the column vectors are also ~ ,and (AB)s = “I +6] “3 given by (AB), =3| 3 +2] | 4 4 5 ot . : 24, The matrix product is AB = 3-3 5 |, and the column’ vectors are also given by (AB), = -16 -6 -8 2 0 -1 2 0 -1 3] 1 |-2] -1 }+2]) 4 |,(AB)o=2}] 1 J+] -1]-—] 4 |, and Ad dels fom] E STE] 2 0 -1 (AB)3 = 1 +0} —1 | + 4 -4 3 1 50 Chapter 2 Linear Combinations and Linear Independence 25, The linear combination ¢;(1 + 2) + c2(a?) = 2a? — 3x — 1 if and only if ce + cya + cpa? = Qa? ~ 8a — 1, These two polynomials will agree for all x if and only if the coefficients of like terms agree. That is, if and only if cy = —1,¢, = —8, and cz = 2, which is not possible. Therefore, the polynomial p(x) can not be written as a linear combination of 1+ and «?, 26, Since ¢1(1 +a) + ¢p(a*) = —23 + 32 +3 has the solution cy = 3,¢, = —1, then p(x) can be written as a linear combination of 1+ and z?, 27. Consider the equation ¢1(1- @) + ¢9(—x) + ¢3(a? +1) + c4(20° ~ a + 3) = o8 — 22 +1, which is equivalent to (er + 63 + ¢4) + (1 — cg — ca) + ca? + 2eqar? After equating the coefficients of like terms, we have the linear system ¢ ~ 0°? which has 2c4 = 1 the unique solution c = 3,c2 = 2,c3 = 0, and cy = 4. Hence, a3 ~ 2@-+1 = $(1 +2) +2(—x) +0(a? +1) + 4 (20° —@ +1) gan" — 6 : . 28. Equating the coefficients of like terms in the equation ¢(1-+2) +¢2(—s) +¢3 (a? +1) +¢4(223 —2 +1) = 28 gives the linear system ¢; + ¢3 + ¢4 = 0,¢1 — cz — cq = 0,3 = 0,2cq = 1, which has the unique solution oO =~}, 02 = —1,c3 =0,c4 = $, Hence a3 = ~4(1 +2) +2 + 4(203 ~ 2 +1), 29, Since , 1 3 lla 131 a 2 7 2/6 |reducestg| 0 1 1) —2a+b |, -l -2 Ojfe 0 0 0] 38a-b+e a all vectors | 6 | such that 3a ~b+¢=0 can be written as a linear combination of the three given vectors. c P 30, Since 1 0 O]a 10 0] a QO 1 0}b -ecluces t 0 1 0} & 01 Ofc |RSS) 0 0 1] ef? 001 0 0 Ofe-b all matrices ‘ ¥ such that ¢ = b can be written as a linear combination of the three given matrices, 31. 32, : V=VitvVe+V3 + V4 V=VitvVatr Vs + v4 = V1 + ve + v3 + (v1 — Qve+ 3v3) : =Vy+ (2v1 _ 4vs) +V¥3 + V4 = 2v1 ~ vo + 4v3 == 38vy — 8v3 + v4 33, Since c; 40, then 34, va = yg Byrn, V=C1V1 bi Henn Cl Cl = CrVa b+ CnVn + OW, +25: + OWm 35, In order to show that $1; = $2, we will show that each is a subset of;the other, Let v € $1, so that there are scalars ¢1,...,¢, such that v = civa +:++ + cKVx. Since c 4 0, then v = @yva fee + £8 (evi); so Vv € Sp. If v € Sg, then v = cyvy +++) + (Cex) VK, 80 V E S1, Therefore $1 = So.