数高代数分-练习题答案

第一§1.(1)(2,

3,2

(4)无交(1)无交 (2)相交直 (3)二平面重 (4)相交直(1)无 (2)(9,3, (3)(0,0, (4)(16,5,3,(1)无

10,1,33

(3)(1,7,3, (4)(−8,1,2,(1)2b1+3b2+3=

(2)任 (3) (4)任2R1↔R2,R2↔R3 (2)1·R2,R3−3R222R1−R3,R1↔R3,1·R12R1↔R2,R4−2R1,R4−5R2,R4+R323.8克、豆46.823.2§ x1=1+x3,x2=1 (2) x1= x1x1 x1确定k(1) (2) (3) (4)−1,λ+

(λ+λ61,λ6−2时，x1λ2x2=λ2,x3λ1时，x11−x2−x3λ−2

λ+2a0,b2时有解，一般解为：x1−2x3x45x5,x232x32x4k=1,(s,−2s,a=7,b=(1)a60

1R1,a

−ER1↔R2,1R1,1R2,R1− (2)R2−3R1,R1−1R2,5R2,,R2+2R1 y=−11x2+13x+ x2+y2+z2−x−y−z=§判断下列各组向量中的β是否可由其余向量线性表出。如果可以，求其表出式(3)β

α1

1α2

1α3

β=−3α1−α2+α31 (4)不15 5(1) (2)6=0且6= (3)(1) (2) (3)不 (4)不βb1b2)Tβk1α1k2α2有唯一解:k12b13b2,k22b23b1(1,−1,0)T,(1,0,根据定义，利用数的加法结合律和乘法对加法的分配律即证α=(a1a2an)Tkα=(ka1ka2kan)T。如kα=0ka1=0,ka20,···,kan0k60时，显然a1a2···an0α0L(α1α2αs§(1)(1,−2, (2)(1,2, (3)只有零 (4)(−2,2,1,(1)s(3,−1,0)T+t(5,0, (1)s(2,7, (2)(1,−1,0)T+s(1,0,(1,−1,0)T+s(1,0,(1,0,1)T+s(2,−2,(1)−210)Ts(5−2 (2)无(3)(0,0,0,1)T+s(−3,2,7,0)T+t(−13,4,0, (4)(−3,3,5,§(1) (2) (3) (4) (5) (6) (1)任 (2) (3)− (1)线性无 (2)线性相 (3)线性相 (4)线性无s是奇数，线性无关；当s是偶数，线性相关只需证明必要性。设有不全为k1k2ksk1α1k2α2···ksαs0i使得，ki60ki+1ki+2···ks0。由此即证。根据定义证明x1β1x2β2x3β30等价于x1α1x2α2x3α30,ax1bx2cx30k(3,1,§等(1){α1,α2,α4};α3=α1−{α1,α2,α5};α3=2α1−α2,α4=α1+(1) (2)

±√n−α1k1β1l1β2,α2k2β1l2β2,α3k3β1l3β2.取不全为零的x1,x2,x3使x1k1x2k2x3k30x1l1x2l2x3l30于是有x1α1x2α2x3α30。(1){α1α2αs}的极大线性无关组可由{β1β2βt}的极大线性无关组线性表显利用(2){α1,α2,...,αs}一个极大线性无关组也是{α1,α2αs},{β1,β2,...,βt}的一个极大线由此得{β1,β2,...,βt}可由{α1,α2,...,αs}线性表出.利用{α1,α2αs}和{β1,β2βt}极大线性无关组即证AA的行数；AA此r{α1,α2,...,αn}=n。(0,−1,0,0)T+k1(−1,−2,1,0)T+(3,2,0,§产出AEMTAEMT0(2)PA=203.28,PE=59.14,PM=132.93,PT=x1=80+x6,x2=−30+x6,x3=40+x6,x4=−80+x6,x5=60+x1160,x250,x3120,x40,x5140,x6≥n(n+1)(2n+6n(n+1)(2n+1)(3n2+3n−第二§(300003000000001.

(

2)

00

00 当n为偶数时5 5

(01

1 cos(nϕ)− (10) { 原矩阵,当n为奇数时用数学归纳法 1 2 −21

−84 0

,a,d,g为变量ABBA对角线上的相应元素相等§提示:A+AT为对称矩阵,A−AT为称矩阵§k不等于零时,kE可逆,1E

−11

0 1adbc6=0时,矩阵可逆,且逆矩阵为

ad− (E−A)−1=E+A+A2+···+Am−1(A−2E)−1=E+A2A−1=−A3+5A−2A2§

— 1122. 3

—

– — 34 (3) −1 1 1−1 ·· ··

·· ··

1(03)21(03)2724.(1)7

10

§

25. A11B12+ A11B13+ A22B23+A23B33 §第三§ (1) (2)(n−1)(n− (3)n− (4)n(n− 2n(n−1)−2 (1)(−1)(n

··· (2)(−

b1b2···(3)5,§

−1)(n a1a2··· (4)(1) (2) (3) (4)[(n−1)b+a](a− (5)(6)an+an−1x+···+ (7)(−b)n−1(a1+a2+···+an− (8)∑(9)a1 n (10)0 (11) (12)xn+(−1)n+1yn7.§(1) (2) (3)(λ−1)2(λ−(λ−1)(λ−(1)(−1)n−1(n−1)! a (2)

(ai−i=1

§

(3)n+ (4)(x−y)(n−2)(x2− (5)−2(n−n(n−1)n+1 2 2 (7)1−a+(−1)a+···+(−1)na(1)(4,2, (2)(1,0,λ1λa6=1b6=C=E+B,|C|=−2(−1)n+1f(x)=x2−+§(1)··a1k?··b1k(3)(a2−6(4)?

.?

?25.

·· akk? ·· 第四§(1) (2) (3)不 (4)不 (5)不§2x4−2ax+ a=25，b=−3，c +f(x)g(x的次数相等而且它们的首项系数之和等于零时，次数公式中的小于号成立；f(x8x96x74x7)3(2x53)7a0x62a1x61···a61xa62f(1)=(8−6+4−7)3(2−3)7=a0+a1+···+a61+a0a1···a61a62−1)3(−1)7证法一由于x51+1=(x−1)(x50−x49+x48−x47+···+x2−x+x51−1=(x+1)(x50+x49+x48+x47+···+x2+x+x1021x21)f(xx1021x21f(x没有证法二f(−xf(x§(1)q(x)=x2−x+4,r(x)=−4x+(2)q(x)=1(3x+11),r(x)=10(x− {m=

{p=m2+(1)p=1—m2,q=

p=1+

q=(xd1)|(xn1ndqr(0rd(xn−1)=xq−1=(xqd−1)xr+xr−所以(xd1)|(xr1r0d|nndq.xn1xd)q1xd1)(x(q−1)dx(q−2)d···由于g1(x)g2(x)|f1(x)f2(x，f1(x)|g1(x)f1(x)f2(x)=g1(x)g2(x)q(x)g1(x)=f1(x)q1(x)将下式代入上f1(x)f2(x)=f1(x)q1(x)g2(x)q(xf1(x)6=0，利用消去律f2(x)=g2(x)|f2(x因为(x21)g(x)=x2n+21，所g(x)=x2n+21.又(x41)f(x)=x4n+41(x2−f(xx4n+41。于是f

x4n+4− x2−

x2(n+1)+ = · = (x4− x2n+2− x2+易见：g(x|f(xx21|x2(n+1)1当且仅当(i2)n+110n(1f(xaxb)q(xrr是一个数.x=brfb 2§2

r=f(−3)=g(x60(f(x)g(xd(x60d(x|f(x),d(x|g(x).d(x)|(f(x)− d(x)|k(xf(x)−h(x)g(xg(xk(x)|f(xk(x)|d(x所以(f(x)g(xf(xh(x)g(x)d(x)=(f(x)g(xd(x也是(f1(x)g1(x显然，d(x)|f1(x)d(x)|g1(xϕ(x)|f1(x)ϕ(x)|g1(x) f(x)

f1(x)ad−

ad−bcg(x) −cf(x) gad−bc ad−bcϕ(x)|f(x)ϕ(x)|g(xϕ(x)|d(xd(x)=(f(x)由于(f(x)g(x1yx有(f(y)g(y1u(y),v(yf(y)u(y)+g(y)v(y)=y=xmf(xm)u(xmg(xm)v(xm1由(f(x)g(x))=1(f(x)h(x))=1u1(x),v1(x),u2(x),v2(x使f(x)u1(x)+g(x)v1(x)= f(x)u2(x)+g(x)v2(x)=两式f(x)[u1(x)u2(x)f(x)+v1(x)u2(x)g(x)+u1(x)v2(x)h(x)]+g(x)h(x)[v1(x)v2(x)]=由互素的充要条件即得利用上题做归纳即得设(f(x)g(xd(x，f(x)=d(x)f1(x),g(x)=d(x)g1(x。这时(f1(x)g1(x))=1(fn(x),gn(x))= 于(fn(x),gn(x))=(dn(x)fn(x),dn(x)f =dn(x)(fn(x),fn(x))=dn(x)=(f(x), =

(f(x), (f(x),((f(x), (f(x),() (f(x),g(x)),(f(x),(

)两边消去(f(x)g(x

f(

)),

(

))=fx), fx),f(xg(x0那么根据定义,f(xg(x的最大公因式0.f(x)g(x)不全为零g(x)60,f(x)g(xq1(x)及))))及余式r2(x，如r2(x)=0则最大公因式为r1(x)，否则用r2(x)r1(x),如此继续下去,因为余式的次数每次降低,所以作了有限次这种除法后,必然得出这样一个余式rs(x),它整除前一个rs−1(x).这样我们得到一串等式:f(x)=g(x)q1(x)+g(x)=r1(x)q2(x)+r1(x)=r2(x)q3(x)+.rs−3(x)=rs−2(x)qs−1(x)+rs−1(x)rs−2(x)=rs−1(x)qs(x)+rs(x)4.1知：rs(xf(x),g(x的一个最大公因式将倒数第二个式子变为：rs(xrs−2(xrs−1(x)qs(x一步步往回带入即可得到多项式u(x),v(x)使得本题的证明给出了如何求两个多项式得最大公因式的方法x+§(=⇒fk(xfk(00，所f(0f(xxa)(xb)q(xAxB f(b)=bA+解方

A=f(a)−f B=af(b)−bf(a)a−(1)q(x)=x3+4x+2,r(x)=q(x)=2x4−4x3+3x2−6x+4,r(x)=

a−(1)f(x)=(x−2)4+6(x−2)3+15(x−2)2+18(x−2)+9f(x)=(x+2)4−8(x+2)3+22(x+2)2−24(x+2)+11f(x)=(x+i)4−2i(x+i)3−(1+i)(x+i)2−5(x+i)+(1+2i)(p(x)g(x 或者p(x)|所以(pm(x)g(x))=(f(x)g(x))=1f(x)|gm(x充分性f(xpm(x)f1(x，p(xp(x)-f1(x，∂(f1(x))>0(f(x),f1(x))=f1(x)6=且对任意的正整数mf(x-fm(xp(x)|f(xp(x)fm(xp(x) §

√ω=−13iωω2ωx2x12x2+x+1|f1(x3)+所以f1(1)+ω2)= f1(1)+ω22=0.由此得：f1(1)=f2(1)=x1|f(xnf(10nω，f(ωnf(1所以xn1|f(1)1,1,···,1. α1α2 假设整系数多项式f(x有整数a，那f(x)=(x−a)g(x).易见，g(x是整系数多项式，则此时0−a,1−a中至少有一个是偶数，所以f(1),f(0)至少有一个是偶数，。21±i,1±234.1(二重根2第34.1(二重根2第五§1.否是是(4)否否是2.3.4.否是否(4)§2因为cos2t=2cos2t1即得反证，不妨设f3(x)=k1f1(xk2f2(x又由于f1(x)f2(x)f3(x中任意两个都不互素 所以f1(x),f2(x)=d(x),∂ >0.所以d(x)是f1(x),f2(x),f3(x)的一个非常..略略

0010 00

0010 01

0001 ;维数301(1)(1,2, (2)(1,1,(1)k2时,dimN(A2η4−210)Tη−720k62时,dimN(A1,η−301(2)k2时,dimR(A2k62时,dimR(A些分量不等于0，.§略

(2)(2,5,−1−1(1)

−1

kα4(k∈11101001000111010010000110···(β,β,...,β)=(α,α,..., 11·· n 11···§Pn,(1)dim(W1W24α1α2β1β2W1W2的一组基;W1W2dim(W1W23α1α2β2W1W2的一组基 1β1=1α11β2, 1

∩

=dim(W1W24α1α2α3β1W1W2的一组基W1W2{αα53119−19((1)L(A,A)∩L(B,B)=

−5

),维数是 L(A1A2L(B1B2L(A1A2B1略略提示：类似于矩阵空间分解为对称矩阵与称矩阵的和的情§§ 是是(6)是略ab1000a010000ab1000a010000ab0000ab1000a010000ab1000a010000ab0000a 0(1)A −1

b0 0,c0

0 0,00

ab .

0 0 00b010·· .00·· .00··n−00··0

§(1)λ1=10,(−1,−2,2)T;λ2=1,(2,0,1)T,(−2,1,λ1=3,(1,−1,2)T;λ2=1,(−2,0,λ1=11,(2,1,2)T;λ2=2,(−1,0,1)T,(−1,2,λ1=2,(1,0,4)T,(1,4,0)T;λ2=−1,(1,0,λ1=2,(0,0,1)T;λ2=1,(−1,−2,λ1=−2,(1,1,1,1)T;λ1=2,(−1,1,0,0)T,(−1,0,1,0)T,(−1,0,0,(1)λ1=1+√3i,(i,1)T;λ2=1−√3i,(−i,λ1=i,1(−1+2i,1−i,1)T;λ2=−i,1(−1−2i,1+i,1)T;λ3=1,(−2,1, ( ( ( ( (1)λ=−1

2;λ=2

α=

1−2

(2048,4095,λ=0,a=2,b=9.12,rA−λE)=3 λ0（n1重）;λαTα（一重 反证法.如果A(k1α1+k2α2)=λ(k1α1+k2α2),则λ=λ1=λ2,§)(1)(

3+51003−3·41− 1+3·5656374 22005− 22005− −22005+ a=5,b=a=3,b=PP−1AP是对角矩阵,P−1APP−1AP)TPTAT(PAλ2(ad)λadbc,其判别式为(a−d)24bc0A有两(1)验证|A|−|A|,TrA−Tr不能确λ=αTβ,αβT的特征值λ（一重）0（n1重）.而rαβT)=1所以属0的特征子空间是n−1维的,即证A的特征值只能是±1而rAErAE)=n,所以dimV−1dimV1=n,所A可对角化R(BA2特征子空间的子空间,R(CA1特征子空间的子空间;由于dimR(B)+dimR(C)=n,所以R(B)+R(B)=Rn,A可对角化,且) )A §(1,0,−1)T,(1,−1,0)T,(−1,1,1)T

32 1 a 1a21 1a23 a11+a12 a21+a22−a11 a22 a23 a31 (1)1 10 3@8 40 A−1(0)=L(4ε1+3ε2−2ε3,ε1+2ε2−ε4);R(A)=L(Aε1,A 基4ε+3ε−2ε,ε+2ε−ε,ε,ε;1 2 4 1042

AεAεε 37. 037. 0−3C,可对角化000@4

A@ A@ 38.特征值及相应特征向量：λ1λ2,1+x;λ3,2+4x+ 01λ=0320§1001001–— –—–66 –66

3 42. (2)

2·9105+ 9105−

4·9105−42·9105+0100100 0 00044.23n+2Aλ2(λ1A2(AE0,A3=0

1 2λ2λ

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@k(k−1) 0000000000000

1050. 0.005 01 种群数量呈周期性

0051.

(2)466545194406 (3)增加, 0.25 2 31(1)λ3−aλ2−abλ 2 31提示：应用根与系数的关系 把|λEL|写成两个行列式,得|λEL|=λ

0,然后把0λ0?00λ?个行列式按第一行展开|λE−L|=λn−a1λn−1−···−(an−1b1···bn−2)λ−(anb1···§(1)x−2 (x−2)2 (x−2)3 (x−2)4(x+3)(x−1) (x+3)2(x−1) (x+3)2(x−1)2 (x+3)2(x−1)3(x+5)(x+2)(x−1) (x+5)(x+2)(x−1)2 (x+5)(x+2)(x−1)3(x+5)2(x+2)(x−1) (x+5)2(x+2)(x−1)2 (x+5)2(x+2)(x−1)3(1)m(x)=(x−3)2(x−4)m(x)=(x−3)2m(x)=(x−3)3m(x)=(x−2)2(x−4)m(x)=(x−2)2(x−4)m(x)=x2m(x)=(x−3)3(x+1)00 10000AkE,Am(xxk1m(x的标准分解式中只有一次因式，故A可对角化。§

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61− √+2221003.k(100)Tl(01−1)T(kl不全为零004.00P,P是正交矩阵,AP=PD,D是对角矩阵,A=PTDP是对称矩阵(1略2由(1)即得3)R(AR(AT1021020240240

0

(2)A(1)2x2+x2+3x2−2xx+4x

1 25x2−7x2+8x2−2x2+2x

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1 1

1 2 3(1)10

√1y1

2 102 x=−√3y+ 10x1=1y1+1y2+√1

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2y2 x2=

y1−√1 x3=−1y1−1y2+1 2 √ √√ √

1+43y1−−1

33y5 33 y5 33

4 x2 −y1 y2+x3 y1 y2+ x1=21y1−1y2−1y3+ x2=1y1+1y2

1y3

1(4)y2+7y2−y2−3y2 4 x= 4 y1 y y 2

2 2

x4=1y1+1y2+1y3+1 a=2

x2 2

+√12 +1 91 2 9 9 9 9 97x2+1x2+1x2+2x1x2−91 2 9 9 9 9 9A=QTg1λ2λn)Q,其中Q为正交矩阵应用§1x=y+y+11(1)−4y