代数芳明著课后习题答案高等教育

—（四）1设f(xabcf(xc)f(x)f(xf(ab)f(af(b)f(x)kx(k为常数f(ab)f(af(b)f(x1或1。证上述命题的充分性显然，下证必要性。若f(x)不是常数，因f(x)f(x))n0，并设f(xic)0fxi) 于是x1c,x2c,...xnc也是f(x)的n个根，再 (x1c)(x2c)...(xnc)x1x2...从而c0，与假 ，即证f(x)是常数f(ab)f(af(b中，令b0f(0)0x00f(x的一个根，从而有f(x)xg(xx2t，得2tg(2t)f(2t)f(xt)f(x)f(t)2f(t)g(2t)g(xkf(x)xg(xf(x)kxf(x)0f(2x)f(xx)f(x)fkf(0)f(0)f(0)f(0)f(0)kk0k1f(x1例 在P[x]中，设g(x)0,h(x)为任意的多项式，试证f(f(x),g(x))(f(x)h(x)g(x),证：由已知，可设f(xg(x))d(x)f(x)q1(x)d(x),g(x)q2(x)df(x)h(x)g(x)[q1(x)h(x)q2(x)]dd(xf(xh(x)g(xg(x)d1(xf(xh(x)g(xg(x)的任意一个公因式，则由多项式的整除性质，可得f(x)q3x)d1x)。这表明d1x)f(x)，从而即d(xf(xh(x)g(xg(x1d(x)(f(x),g(x))(f(x)h(x)g(x)例 设f1(x),f2(x),g1(x),g2(x)是实系数多项式，(x21)f1(x)(x1)g1(x)(x2)g2(x)(x21)f2(x)(x1)g1(x)(x2)g2(x)证由(x1)1)x1)2)

(x21)[(x1)f1(x)(x1)f2(x)]6xg2再由(x2)(1)x22，同理可证(x21)|g例 试问：2是否为一元多项f(x)3x44x3x24xg(x)x56x411x32x212x 2是否为f(x)或者g(x)的根时，可采用综合除法，得f(2)800g(2)02f(x2g(x的根。进一步2是g(x)的几重根。g(xg(x)x56x411x32x212xg/(x)5x424x333x24xg//(x)20x372x266xg///(x)60x2144xg(2)g/(2)g//(2)0,g///(2)182g(x例 设有一个三阶行列 f(x1,x2,x3) x1 f(x,x,x)x3x3x33xx 1x1x22x1x2x1x3x23x1x2

12则所求多项式f中相应的初等对称多项式i(i1,2,3) 的的 3 1 f(x,x,x)x3x3x33xxx3a 12 x10x21,x31,bf33 1 设,,分别为方程x3ax2bxc0的三个根，则333()33()()a33abb33abc3c333cy3(a33ab3c)y2(b33abc3c2)yc3二 g(xf(xq(xr(x1）f(x)x33x2x1,g(x)3x22x2）f(x)x42x5,g(x)x2x解1）q(x)1x7r(x)26x 2）同理可得q(x)x2x1,r(x)5x1）x2mx1|x3px2）x2mx1|x4px2解(p1m2)x(qm)p1m2 qmx2mx1|x3pxm(2pm2)q1pm2于是当m0时，代入（2）pq1；而当2pm20时，代入（2）q1。m q pq 或pm2 x2mx1|x4px2g(xf(xq(xr(x1）f(x)2x5x5x38,g()x32）f(x)x3解1）

2,x()x1q(x)2x46x313x239xr(x)q(x)x22ix(5r(x)9把f(x)表示成xx0 cc(xx)c(xx)2...c(xx)n 1）f(x)x5,x02）f(x)x42x23,x0f(x)x42ix3(1i)x23x7i,x01）f(x)15(x1)10(x1)210(x1)35(x1)4(xx42x231124(x2)22(x2)28(x2)3(xx42ix3(1i)x23x(7(75i)5(xi)(1i)(xi)22i(xi)3(xf(xg(x1）f(x)x4x33x24x1,g(x)x3x2x2）f(x)x44x31,g(x)x33x23）f(x)x410x21,g(x)x442x36x242x解1）f(xg(xx2）(f(x),g(x))3）(f(x),g(x))x222x求u(xv(x使u(x)f(xv(x)g(x)(f(x1）f(x)x42x3x24x2,g(x)x4x3x22x2）f(x)4x42x316x25x9,g(x)2x3x25x3）f(x)x4x34x24x1,g(x)x2x1）(f(x),g(x))x22r2f(x)q1(x)g(x)g(x)q(x)r(x)r r2(x)g(x)q2(x)r1(x)g(x)q2(x)[f(x)q1u(x)q2(x)xv(x)1q1(x)q2(x)11i(x1)x(f(x),g(x))x且u(x)1x1,v(x)2x22x 由f(xg(x))1u(x)x1,v(x)x3x23xf(xx31t)x22x2ug(x)x3tx2u式，求t,u的值。 f(x)q1(x)g(x)r1(x)(x3tx2u)(x22xu)g(x)q2(x)r1(x)r2(x)(x(t2))(x22xu)(u2t4)xu(3(u2t4) u(3t)u1 u2t t g(x证d(xf(xg(x)的公因式。另设(x)f(xg(x)(x)|d(xd(xf(xg(xs(x与t(xd(x)s(x)f(x)从而由(x)|f(x),(x)|g(x)可得(x)|d(x９．证明：(f(x)h(x),g(x)h(x))(f(x),g(x))h(x) (h(x)的首系数为１。证因为存在多项式u(x),v(x)使(f(x),g(x))u(x)f(x)(f(x),g(x))h(x)u(x)f(x)h(x)上式说明(f(xg(x))h(x是f(x)h(x)g(x)h(x)另一方面，由(f(xg(x|f(x(f(x),g(x))h(x)|f(f(x),g(x))h(x)|从而(f(x),g(x))h(x)是f(x)h(x)与 的一个最大公因式，又因f(xg(x))h(x(f(x)h(x),g(x)h(x))(f(x),f(xg(x f

, (f(x),g(x))(f(x),g(x))证存在u(xv(x(f(x),g(x))u(x)f(x)f(xg(x(f(x),g(x))1 f (f(x), (f(x), f

, (f(x),g(x))(f(x),g(x))f(xg(xu(x)f(x)v(x)g(x)(f(x),那么(u(xv(x))1证明：如果f(xg(x1,f(xh(x1，那么(f(x),g(x)h(x))1证由假设，存在u1xv1x及u2xv2xu1(x)f(x)v1(x)g(x)u2(x)f(x)v2(x)h(x)（1（2）

[u1(x)u2(x)f(x)v1(x)u2(x)g(x)u1(x)v2(x)h(x)]f[v1(x)v2(x)]g(x)h(x)(f(x),g(x)h(x))f1x),...,fm(xg1xgn(x(fi(x),gj(x)) (i1,2,...,m;j1,2,...,(f1(x)f2(x)...fm(x),g1(x)g2(x)...gn(x))(f1(x),g1(x))(f1(x),g2(x))(f1(x),gn(x))(f1(x),g1(x)g2(x)...gn(x))(f2(x),g1(x)g2(x)...gn(x))(fm(x),g1(x)g2(x)...gn(x))(f1(x)f2(x)...fm(x),g1(x)g2(x)...gn(x))证明：如果(f(xg(x))1，那么(f(x)g(x),f(xg(x))1证由题设知(f(xg(x))1，所以存在u(xv(xu(x)f(x)v(x)g(x)u(x)f(x)v(x)f(x)v(x)f(x)v(x)g(x)即[u(x)v(x)]f(x)v(x)[f(x)g(x)](f(x),f(x)g(x))(g(x),f(x)g(x))(f(x)g(x),f(x)g(x))f(x)x32x22x1,g(x)x4x32x2x解(f(x),g(x))x2x所以它们的公共根 21）f(x)x55x47x32x24x2）f(x)x44x24xf(x)5x420x321x24x解(f(x),f(x))(xf(xx22）f(x)4x38x4(f(x),f(x))1f(x求t值，使f(x)x33x2tx1有重根。 f(x)有三重根x1时，t3。若令x33x2tx1(xa)2(x32ata2 1（1（3）2a33a21a1,a1,a 1，得t3,t3,t 当t1,23f(x3x1；当t34f(x2x2解令f(x)x3pxq，则f(x)3x2p0q0,f(xx3p0af(x的重根，那么af(xf(xa3paq3a2p由（1）可得a(a2pq，再由（2）a2p3a(pp)3a24

a2p3p4p327q2综上所叙即知，当4p327q20x3pxq如果(x1)2|ax4bx21af(x)ax4bx2f(x)4ax3由题设知，1f(xf(x的根ab14a2b 证明：1x 不能有重根 f(xf(x)1x1x2...

f(x

f(x1xn(f(x),f(x))(f(x)1xn,f(x))(1xn,f(x)) f(xg(xxaf(xf(af(xf(ak+32证g(x)xaf(x)1[f(x)f g(x)x2

f由于f(xk重根，故g(xk1重根。代入验算知g(x现在设g(xs重根，则g(xs1g(xs-2重根。s2k1skx0f(x的kf(xf(xf(k1)x0f(kx x0f(xk重根,f(xk1f(xk2f(k2)xxf(kx的根。于是 f(x0)f(x0)...

k)x0,f(k)(x 充分性：由f(k1)x0，而f(kx0，知x是f(k1)x)的一重根。又由于 举例说明段语“f(xm重根，那么f(xm1f(x)

xm1f(xxm0为m0f(x证明：如果(x1)|f(xn，那么(xn1)|f(xn证要证明(xn1)|f(xnf(1)0（xn看作为一个变由题设由(x1)|f(xn证明：如果(x2x1)|f1(x3xf2(x3(x1)|f1(x),(x1)|f2x2x1的两个根为和2cos2isin 所以和2f1(x3xf2(x3的根，且31f(1)2f(1) 解在复数范围内xn1(x1)(x)(x2)...(xcos2isin jnj(0jxn1(x1)[x2(n1)x1][x2(2n2)x1]...[x2

2)xjnjjj2cos2j(j1,2,...,n1 xn1(x1)(x1)[x2(n1)x1][x2(2n2)x1]...[x2( 2)x1）x36x215x2）4x47x25x3）x5x46x314x211x解有一个有理根：2 有两个有理根： , （即有2重有理根 1）x22）x48x312x23）x6x3xppx1,px44kx1,k1）因为1x21命题f(xxy1xy1gyfy1)gyfy则f(x)gy)或者同时可约，或者同时不可约。f(xf(xf1xf2x，从而g(y)f(y1)f1(y1)f2(yx6x31xy1(y1)6(y1)31y66y515y421y318y29yf(xxppx1xy1g(y)f(y ypC1yp1C2yp2...Cp2y2(Cp1p) pp|Ci(i1,2,...,ppp2|pgyf(x也在有理f(x)x44kxxy1g(y)f(y1)y44y36y2(4k4)y4kx2xxx2x2xxx2x2xxx1 1 1 1 2(xx)2(xx)2(xx 1 1 1 x2x2x2x2x2x2x2x21 1 1 1解1）对称多项式的首项为x2x，其为(2,1,0)，1 1 1x2xxx2x2xxx2x2xxx23xx1 1 2x2xxx2x2xxx2x2xxx22xx1 14）(x22xxx2x22xxx2x22xxx2 1 1 2 1x4x211由此可知多项式时六次对称多项式，且首项为x4x2，所以 1对应 1 112 3原式=22a3b3cd 1 1 12 x10x2x30,则原式左边012,21,3代入（1）式，得b4x1x21,x32，得dx1x21,x31acx1x2x31,3ac2（原式2243431827 1 12 14）原式=x2x2x22x 2x2x2x2x2x2x1对应 2 1 设原式2a 1 x1x2x31,x40,a2x1x2x3x41,得b2。原式22 1 原式x2x2x2x3xxxx3xxxx31 12 12 12 1 12(x2x2x2x2x2x 1 12

x3xxxx3xxxx32212 12 22 1 21 x2x2x2x2x221 原式222221 1 2 原式x2x2x22(x2x2xx2xx2xx2x31 1 12 1 (x2x2x2x2x2x23x2xx3xx2x3xxx21 2(x2xxx2x2xx2xxx2xx2)2xx1 12(x2x2xx2xx2xx2x3)21 12 1 2 x22x2x2... x22 x2xxx2...xx21 1 2 1 原式2221 1 2 2） 2x2 1 1 11 1 1对应 1 21设原式4a2b2c 1 令x11x21,x3x4xn0,得b2x1x21,x3xn0a4x1x2x31,x4xn0,得c4x1x21,x3x41,x5xn0,d4。原式442224 1 1 原式134原式22 1 原式244159设a1a2a3是方程5x36x27x30(a2aaa2)(a2aaa2)(a2aaa2 1 2 1 解1a1a26

7, (a2aaa2)(a2aaa2)(a2aaa2 1 2 1 22331 1 2a39aa27a 1 证设原方程的三个根为1,2,3(2)(2)(2)239 1 2a39aa27a 1 f1(xaf(xbg(xg1(xcf(xdg(xadbc0(f(x),g(x))(f1(x),g1证d(x)(f(x),d(x)|f1(x),d(x)|g1其次，设(x)f1(xg2(x的任一公因式，只需证明(x|df1xaf(xbg(xg1xcf(xf(x)

f(x)

g adbc adbcg(x) f(x) g

adbc adbc|f1,|g1|f,|从而(x|d(xd(xf1xg1xf (f(x),g(x))(f(x),u(x)f(x)v(x)g(x)(f(x),的u(x)与v(x)(u(x)) ,(v(x)) f (f(x),g(x)) (f(x),g( 证存在多项式u1(xv1(xu1(x)f(x)v1(x)g(x)(f(x),u f

v

(f(x),

(f(x),(u(x)) (f(x),g(x)) 则(v(x)) f (f(x),g(x)) (v(x)) f (f(x),g(x)) g f (f(x),g(x)) (f(x),g( f (f(x),g(x)) (f(x),g(x)) f 这样（1）式左端的次数f(xg(xf(xg(x0 (u(x)) ,(v(x)) (f(x), (f(x), 此时u1(xv1(x若(u(x)) (f(x),g(x)) 则 除u1(x)，可(f(x),u1(x) (f(x),

(r(x)) (f(x),g(x)) u1(x) (f(x), (f(x),

(f(x),。再将u1xs(xf(xg(x))r(x代入（1） f(x) f(x) v(x)] (f(x),g(x)) (f(x),g(x)) (f(x),g(x))令u(x)r(x),v(x) f(f(x),

f(xg(xf(xmg(xm也互素。证由假设，存在u(x)和v(x)使u(x)f(x)v(x)g(x)u(xm)f(xm)v(xm)g(xm)),f1(x),f2(x),...fs1(x),fsf1x),f2xfs1x),fs(x(f1(x),f2(x),...fs1(x),fs(x))((f1(x),f2(x),...fs1(x)),fsd1x)(f1x),f2xfs1x))于是d1xfs(xd(x)(d1(x),fs则d(x)|fi 若设(xf1x),f2xfs1x),fs(x(x)|这样(x)d1(xfs(x的一个公因式，又可得(x|d(xd(x)(f1(x),f2(x),...fs1(x),fs)f1xf2xfs1xd1x)成立，再证命题对s也成立。事实上，存在p(xq(x)，使d(x)(d1(x),fn(x))p(x)d1(x)q(x)fsp(x)[v1(x)f1(x)v2(x)f2(x)...vs1(x)fs1q(x)fs令ui(x)p(x)vius(x)

1）f(x)|m(x),g(x)|m(x)我们以f(xg(x1f(xg(x的首项系数都是1，那么[f(x),g(x)] f(f(x),证(f(x),g(x))d则f(x)f1(x)d(x)，g(x)g1(x)df(x)g(x) (f(x),

f(x)g1(x)g(x)即f(x)

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

，g(x)

f(x)g(x)(f(x),M(xf(xg(x)

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

|M(xM(x)f(x)s(x)即f1(x)d(x)s(x)f(x)s(x)g(x)t(x)g1(x)df1(x)s(x)g1x|f1x)s(x。由于f1xg1x1g1x|s(xs(xg1M(x)f(x)s(x)f(x)g1(x)q(x) (f(x),

f(x)g(x)|M(f(x),证明：设p(x)是次数大于零的多项式，如果对于任何多项式f(xg(x)，由p(x|f(x)g(x)p(x|f(xp(x|证p(xp(x)p1(x)|p2p(x|p1xp(x|p2证明：次数0且手项系数为1的多项式f(x)是一个不可约多项式的 条件为：对任意的多项式g(x)必有(f(x),g(x))1，或者对某一正整数m,f(x)|gm(x)。证必要性：设f(x)ps(x)（其中p(x)是不可约多项式则对任意多项式g(x)，有1）(p(x),g(x))1；或2）p(x)|1）有f(xg(x))12）ps(x|gs(xf(x|gs(xms，即证必要性。充分性：设f(x)不是某一个多项式的，则f(x)p1(x)p2(x)...pn g(x)p1(xf(xg(xf(xg(x1f(x|gm(x（m为某一正整数）证明：次数01f(x是某一不可约多项式的的充分必要g(xh(xf(x|g(x)h(xf(x|g(x，或者对某一正整数m,f(x|hm(x。f(x|g(x)h(xh(x1）f(xh(x))1f(x|g(x2）f(x|hm(x)(m为某一正整数。f(xg(x))1或f(xg(x))d(x)f(x)f1(x)d(xf(x)|f(x|f1(xf(x|gm(xm7题的充证设f(x)xnaxnmb，则f(x)xnm1[nxm(nf(xg(x)nxmnm)ag(x)m个根都是单根，因而f(x)没有不为零且重数大于2的根。f(x|f(xnf(x证设af(xf(x|f(xnaf(x|f(xnf(xn)a,an,an2f(xf(xmf(x最多只可能有mkankan(ankn1)f(x|f(xf(xnnf(x 设a1,a2,...,as是f(x)的s个不同的根且它们的重数分别为1,2,...,s由于f(x)n1(11)(21)...(s1)n1sns1f(xa1，且重数为1n1。故f(x)n重根。设a1a2an是n F (xa)F(a i 2）f(xF(xn f(ai)Fni1(xai)F(ai证1）g(x) F(xa)F(a 则(g(x))n但g(a1)g(a2)...g(an)g(x)1nn

F i1(xai)F(ai2）f(xF(xf(x)q(x)F(x (r(x)0或(r(xnr(x0时，结论显然成立。当(r(xn1k(x) f(ai)F(xa)F(ai 则(k(x))n1r(ai)f(ai)k(ai r(x)k(x) f(ai)F(xa)F(a L(x) (xa)F(a

iL(ai)

利用上面的：一个次数4f(xf(2)3,f(3)1,f(4)0,f(5)f(xx0,处与函数sinx2f(xf(0)1,f(1)2,f(2)5,f(3)解1）F(x)x2)(x3)(x4)(x5)f(2)3,f(3)1,f(4)0,f(5)f(x)3(x2)(x3)(x4)(x5)(1)(x2)(x3)(x4)(x 0(x2)(x3)(x4)(x5)2(x2)(x3)(x4)(x 2x317x2203x f(sin00f(0)，sin1f(

，sin0f(F(xx(x)(x2f(x)

F(xx(x1)(x2)(x3)f(x)x2f(xf(0)f(1f(x不能有整证设af(xf(x)(xa)f(0)f(1)(1a)f 又因为a与1a中必有一个为偶数，从而f(0)与f(1)中至少有一个为偶数，与题设 故f(x)无整数根。xnaxn1...a f(x,...,x)g(/,..,/ 其中(i1,2n1)x,.,x /x x 1x/ x 1其中i为x1,x2 1

2

将（3）代入（2）可知，/是x,a,a,..., /p(x,a,a,..., 设令

f(x)(xx1)(xx2)...(xxnxn1xn1...sxkxk... (k0,1, xk1f(x)(sxksxk1... xs)f(x) k g(x)的次数ng(x)0s s1)kk （对1kn 1k 2k k1 sk1sk12sk2...(1)nnskn证1）

（对knf(x)

fi1xxk1f(x) xk1fi1x nxk1xk f f x i1xn

(xkxxk1...xk)f(x) g(x)i1x

f是一个次数nxk1f(x)(sxksxk1... xs)f(x)

k f(x)xn1xn1...(1)n(sxksxk1... xs)(xnxn1...(1)n) k 当knxn的系数，首先由于(g(x))ng(xxn的项，所以等式左端含xn的系数为s s(1)ks 1k 2k k1 0s s(1)ks(1)k(nk 1k 2k k1 0 s s(1)k 1k 2k k1 当kn时，等式(nxn0xn的项的sk1sk1...(1)nnsksk1sk1...(1)nnskn解1）当n6时，由上题可得 s22 s33 1 s442422 1 1 s553525255 1 1 1 1 2 s66463922622 1 1 1 1 326612 2 1 12 s664639226212 1 1 1 1 12623631 2 s553525255 1 1 1 1 2s44242 1 1 s33 1s7s5i 5证这时n6s110s1330，所以30即而s71s62s53s44s35s56故s7s5i 2 5求一个n次方程使s1s2sn10解设此方程为xn1xn1...(1)nnxn1)nn0xna求一个ns1s2..snxn1xn1...(1)nn k11 即

(k2,3,..., k

12，13，...，1 2 3! n! 2 x1

1

第二 3478269 1798635 8765432解:1) 01133011 10454301所以此排列为偶排列 876543219912所以此排列为偶排列选择i与k1274i56k9成偶排列1i25k4897排列解 1)当i8,k3时,所求排列的逆序数为 00413110故当i8k3时的排列为偶排列2当i3,k6时, 01011011故当i3,k6时的排列为奇排列解 决定排列nn1"21(行列式第1解 因为1与其它数构成n1个逆序,2与其它数构成n2个逆序nn1"21n1n2"22故当n4k,4k1时，排列为偶排列；当n4k2,4k3时排列为奇排列。x1x2xn1xn的逆序数为kxnxn1x2x1的逆序数是多解 因为比xi大的数有nxi个,所以12"n1nn2x1x2xn1xn的逆序数为kxnxn1x2x1 k2

a32a43a14a51a66a25解 (1)2345163126451441解：所求的各项应是a11a23a32a44，a12a23a34a41，a14a23a31a4200"01010"000"20002"0##%#####%#.0n"00000"nn0"00n00"00"0100#"%2#0#0#.n"0000"00n解：1）所给行列式的展开式中只含有一个非零项a1na2,n1"a，

n( n!123"n11n1a1,n1a2n2"an1,1ann，

n d

解：行列式展开的一般表示为a1ja2ja3ja4j 因此原行列式值为0.x121x1x121x132x1111x由

11"11"1## 11"而行列式的值为0，这说明带正号与带负号的项的项数相为偶排列时，该项前面所带的符号为正，否则为负号. x

22 22

aa2 a2

其中a1 是互不相同的数.1）由行列式定义，说明Px是一个n1次多项式；xn1的对应项的 a

a a

a a

a

3）若用a1a2,"an1x时，则由行列式的性质知根a1a2,"an1.又因为Px是一个n1次的多项式，所以a1a2"an1必是Px的全部根. 1）

x2） x x

11111111111111111 6）bcd

abc

bc

abcd 1解：1）原式= 44310501=105 4431050 2x2 x

327294105 x原式=2x2 x 2x2

2(xy)0

x =2(x

xx

y2x3y3 x

6

6

68

=20

20 4160xx0xx00x0001111110 1 2a2a2a2a2b2b2c2c2c2d2d2dcd=0b c a

a2a22a222b222c222d22cd b2

c2

a2

2a1 b1 ab c a左边

a2b2

c2

a2ab=2a1b1c1a2b2

abcc1121401214021002100031122）21014解A116，A120，A130，A14A2112，A226，A230，A24A3115,A326,A333,A340A417,A420,A431,A442）A117,A1212,A13A216,A224,A23A315,A325,A335

111212 111212 12254321 12254321

12）13

2331211021233121102122103523）

4）

5

1111011511110115002000

2）112222211162041201201433=- 12 =-12463654332 212142121435241 原式=

10 2121106212110606083

13 6

4832102210216042112012040132480128101242601206 =1843142 300302602=-

5

8

0=-

123810 计算下列nxy0"000xy"00a1a1"a1#0#x#"%0#0#a2#a2#"%a2#000"##anan"any00"0xxx"xxx"x2"##%#xx"xn 2 4） 3 # 2 1123"nn10"0002"00###%##000"20000"n1a1a1"a2"##%#an"当n3时，原式；当n1a1b1 3）原式=

m

x2m xn

m

nxm 1 21 04）原式= 1

1 0 1 0 2 # # # 100 n2n2！5）各列加到第1列得到

100

nn20230""n0n002"00###%##000"0000"nn=1n11n12111"110"01）0"0###%#100" n1aa1a2"a0 a ix00x00"0x0"00x"0##%%## x

xn xn1"axa

n1 00"001"0001%00##%%%#000%000"110"0010"00121"00012%00##%%%#000%21000"121111"11111"11111%11##%%%#111%11111"11

n1"1a1 a

i证：4）分别将第i(i,"n1)行乘以-证

n" "ia0i

n=a1a2"(a0

)i1a0000"000"000"0###%#a0000"000"000"0###%#000"0000"xn1

n2" 12xn212

n3"

ax

#xx x x1n1xn xn1"axa 1n1xn xn1"axa xn xn1"ax DnDn1DnDn1Dn1Dn2此式对一切n

Dn

2

Dn3

2n222DnDn1所 从 Dn

n1 2cos21cos2D 22Dn2cosDn1Dn2cosn111"1111"11011"11011%11##%%%#011%11111"11 n

1

#0##0#%%%0%0#0000""0000%0###%# = n1aa1a2"1 a i用法则解下列方程2x1x23x32x4 x12x23x32x41）3x3x3x2x 2）2xx2x3x 3xxx2x

3x2xx2x 3x1x23x3x4 2x13x22x3x4x12x22x34x4x5 2xx3x4x2x

5x16x2x5x6x 3）3x1x2x32x42x524）x25x36x44x3x4x2x2x x1x2x32x43x5

x5x6x x45x5解：1）d70,d170,d270d370d470xd11,

d

d41 2）d324,d1324,d2648,d3324,d4648xd11,

d

2,

d

2 d24,d196,d2336,d396,d4168,d5312xd14,

d

14,

4,

d

7,

13

4）d665,d11507,d21145,d3703,d4395,d5212x1507,

229,

37,

79,

212fxccxcx2" 使 i,,",ccaca2" an1 2

n1 c " an1 1 2 n1 .................ccaca2" an1 1 2

n1 这是一个关于c0c1"cn1设水银密度h与温度thaatat2at tDC0h求t15，40（准确到两位数）.解：将th的实验数据代入关系式haatat2at 得a13.60 10a1 a320a400a8000a d12106d150000,d21800,d3由法则可求a10.0042,a20.00015,a3h13.600.0042t0.00015t2 t再将t15,t40ht15 ht40第三章 x2x3x2xx3x2x2xx

x1x3xx53xx2xxxxx14x2x3x4x5

2x3x4x5x2x

x2xxxx xxx

2x3x3x2x3) 4)

x3xx 4x11x13x16x 2xxxx 3x1

3x2xxx 6)2x3xxx 2xxx3x 5x1x2x322xxx3x 5x5x2x 解1） 1 3 5 4 0 1

1 1 3 2 1 3 21 0 1 0 2 1 0 1

1 000 0 00001 0 0000 0 00001 0 000

000rank(A)rank(B)4x1x412xx 2x

x11xx3x3x 1 2 0 3 2 1

1 2 0 3 2 1 7 5 16 22021 2021100383 08307 00003rank(A)4rank(A) 1 2 3 4 4

1 2 3 4 4 1 1 0 1 2 2

12 rank(A)rank(A)

x2x 7 9 9 9 20 0 x17x28x39x417x19x20x x k 9 x k20 17 17x

x3 2 1 1 1 1

2 1 1 1 1 1

2 1 1 1 1 2 rank(A)4rank(A)所以原方程组无解 1 2 2

2 0 2 0 0

0 0 0 0 0

1 1

5x27x3 5x3x4 x2 k kx

16

(0,0,解1）k11k22k33

kkk kkkk k5 k1 k1 k 5111

4证明：如果向量组1,2,",r线性无关，而1,2,",r,线性相关，则向量可由1,2,",r线性表出.证k11k22"krrkr1k11k22"krr成立，这与1,2,",r线性无关的假 ，即证kr10.r

kiikri可由1,2,",ri(i1,i2,",in)(i1,,",n)，证明：如果1,2,",n

0证

k11k22"knn11 21 n1k11 21 n1kk"k12 22 n2ij0，故齐次线性方程组只有零解，从而1,2,",n设t1t2,"trrnii(1,ti,"tn1)(i1,,"i证k11k22"krr则k1k2"krt1k1t2k2"trkr 式为一个范行列式，即11""11""122"2##%# t t t

(tt)i所以方程组有惟一的零解，这就是说1,2,",r当rn(1,t,t2,"t 1 2 2 rr

则由上面1)的证明可知12,"r是线性无关的.而1,2,",r12,"r延长的向量，所以1,2,",r则k1k3kk k2k3kk1k2k3已知1,2,",sr，证明：1,2,",sr个线性无证设i1,i2,",ir是1,2,",sr个线性无关向量组，如果能够证明任意一个向量j(j1,,"s)都可由i1,i2,",ir线性表事实上，向量组i1,i2,",ir,j大于r， .这说明j可由i1,i2,",ir线性表出，再由j的任意性，设1,2,",s的秩为r，i,i 是1,2,",s中的r 向量，使得1,2,",s中每个向量都可被它们线性表出，证明：i,i,",i是1,2,",s 由题设知i,i 与1,2,",s等价，所 i,i 的秩与1,2,",s的秩相等，且等于r i,i,",i线性无关，故而i,i 是1,2,",s的一个极 证将所给向量组用（Ⅰ）表示，它的一个线性无关向量组用中去，得到向量组（Ⅲ）且向量组（Ⅲ）是线性无关的.证1）由于1,2的对应分量不成比例，因而1,2线性无关.2）因为3312，且由

k1k2k40，因而该齐次线性方程组存在非零解，即1,2,4,5线性相关，所以5注此题也可将1,2,4,5排成54的矩阵，再通过列初等变换

，

2解1）A2

2 3 4

4A 证由题设，向量组（Ⅰ）的极大线性无关组也可由向量组（Ⅱ）的设1,2,",n是一组维向量，已知单位向量1,2,",n可被它们线性表出，证明：1,2,",n线性无关. 设1,2,",n的秩为rn，而1,2,",n的秩为n

nrn.故1,2,",n设1,2,",nn维向量，证明：1,2,",n线性无关的充分必要条件是任一n维向量都可被它们线性表出. 必要性.设1,2,",n线性无关，但是n1个n维向1,2,",n,必线性相关，于是对任意n维向量，它必可由1,2,",n充分性.任意n维向量可由1,2,",n线性表出，特别单位向量1,2,",n可由1,2,",n线性表出，于是由上题结果，即证1,2,",n11 12 1n axax"axbaxa11 12 1n 21 22 2n an1x1an2x2"annxn对任何b1b2"bn证充分性.由来姆法则即证.i(1i,2i,",ni)(i1,,",(b1,b2,

0x11x22"都可由线性1,2,",n表出，因此由上题结果可知1,2,",n线性无关.k12k22"knnAaij0已知1,2,",r与1,2,",r,r1,",s有相同的秩，证明：与1,2,",r,r1,",s等价.证由于1,2,",r与1,2,",r,r1,",s它们的极大线性无关组所含向量个数必定相等.这样1,2,",r线性无关组也必为1,2,",r,r1,",s 设123"r,213"rr12"r12,"r与1,2,",r 只要证明两向量组等价即可.由题设，知1,2,"r可1,2,",r1(")"2r

1 r1

1r1

"(r

"1r1即证1,2

(i1,,"也可由12,"r线性表出，从而向量组1,2,"r与1,2,",r 2 1） 2） 1

6 5） 0 解1）讨论abx1x2x2 (3)x1x2x21）xxx 2）x(1)xxx1xx3 ax1x2x2x1bx2x3x2bxx 解1）

D 1(1)2( 所以当1x1x2x2

x11k1k2xk 当2当1且2x x ( 3 D 1 33

当0，且1 33215

312

433212 D 1b(a D0a1且b0x2b b(ax2 x12ab b(aD02o若a1A A

4 b 1 02b 01所以当a1,b 2a1,b12

x12kx2x1x2x3x4x503x2xxxx0

x1x23x4x5xx2xx 1） 2x2x6x

2x6x3x4x 5x4x3x3xx

2x4x2x4x7x x12x2x3x4x502xxxxx

x12x2x3x4x502xxx2x3x0 x3）x

7x5x5x5x

4）x2xxx2x3 解1） 1 1 1 6 6 0 x1x2x3x4x5x2x2x6x

x31,x4x5x11,x3x5x51,x3x4

0

1

1

x1x2x4x52x2x2xx0 0 x0,x1，得(7 2

5 6

0 0

1 05x23x33x4x56x9x

2131( , , 312

1

1 22

2

1 1 0

0 x12x2x3x42x5xx 8x4x5x x0,x1，得 (,0, 3x4x5x7xx3x2x2xx

2x13x23x32x4 x2xxxx

4x11123x4xx4xxxx

7x12x2x33xx2xxxx x12x23x3x43x2xxx6)2x136)2x13x2x3x42x2x2xx 解1）rank(A)rank(Ab)4x1x42xx55

1 1 2 x30k0k0 4 rank(A)2x17x28x39x417x19x20x x1,x0x3x x0,x1x13x (3,19,1,0)

(13,20, 17 3 13 17x2 17 20xkkk19k x 3 1 2x x 5 1rank(Ab)35x27x3 5x3x4 ,x4

6,1, ,1,

0 ,5

1)5) 0 1x 2 72xk

5

5x3 0 14 1 6x 5 53x2xxx3x x2x2x6x 解 1 aA 3 b 1 1 1 1 1 1 3 b 于是，只有a0且b22，此时原方程组有解；当a0且b2时，原方程组都无解.a0b2x1x2x3x4x5x2x2x6x

x12k3k4x32k2kx 设x1x2xxx3x4x3x4xx 5ai证0001000100010001000100A010001 1 2 2a a3 4 1

5ai5 5ai5因此，原方程组有解的充分必要条件是ai5其次，当ai0

x2x3xx

x1a1x2a3a4xaaax3a3x3a3a4xa 证1,2,",ra1a2,"ar价，则ai(i1,,"r可由1,2,",r线性表出，从而ai(i1,",r也a1a2,"ar线性无关，且1,2,",ra1a2,"ara1a2,"ar线性表出，即证a1a2"ar也是方程组的一个基础解系.11 21 n1x11 21 n1xx"x12 22 n2证由于方程组的系数矩阵的秩为r，所以它的基础解系所含线性无关解向量的个数为nr.设1,2,",nr是方程组的一个基础解系，1,2,",nr是方程组的任意nr个线性无关的解向量，则向量组1,2,",nr,1,2的秩仍为nr，且1,2,",nr是它的一个极大线性无关组，同理1,2,",nr也是它的一个极大线性无关组，所以1,2,",nr1,2,",nr证明：如果1,2,",tu11u22"（其中u1u2"ut1）证i1x1i2x2"inxnbi(i1,,"由题设，xjx),"xjj1,2,",t是该方程组的t uu"u

ux(k),

ux(k)"ux(k)1 2

t kk

kk

kk a(ux(k))a(ux(k))"a(ux( kk

i kk

kkk

u(ax(k)ax(k)"ax(k) i1 i2 in ukbibiukbi(i1,"k k所以u11u22"utt

2x33x2x2x4x23x在解f(xg(x220000220000220R(f,g)00022100001000010(3)(34228故当3R(f,g)(3)(34228157)f(xg(xRfg0还可求得3它们皆可使f(x)