高等代数与解析几何 习题解答

1?n?“?e?;?/ 1?“? 1.e?“?5/: (1) + 1 1 1 ! ;(2) 1 1 12 2 + 1 ! ; (3) 12 1 3 12+ 232 1 + 122+ 1 ; (4) 22 132 2 2+ 3 22 3 2+ 2+ 22+ 2 ; (5) + 200 1 + 20 01 + 2 ; (6) 00( 1) 02 10 ( 1)200 . ): (1) diag(1, 1). (2) diag( 1,( 1)( 2). (3) diag(1,0). (4) diag(1,( + 1),( + 1)2 ? 1 2 ? ). (5) diag(1,1,( + 2)3). (6) diag( 1,( 1)( + 1),( 1)2( + 1). 2.?e?“?d: 180 1?n?“?e?;?/ (1) A= 32 4 + 3 2 22 52 4 + 3 2 2( 2)2 , B= 2 3 + 32 3 3 2 2 + 14 72 5 2 3 + 22 4 2 . (2) A = 2 22 1 + 1 0 + 11 ( + 1)22+ + 1 , B = 122+ 1 1 22+ 1 + 1 . ): (1)?d; (2)?d. 3.e?“?,=?_?e?_?_. (1) + 1 1 + 3 + 1 ! ;(2) 2 22 + 2 + 1 ! ; (3) 1 2 + 2 + 12 1 + 2+ 1 ; (4) 12 2+ 122 1 . ): (1)?_,_? 1 4 + 1 + 1 3 + 1 ! . (2)?_,_?1 2 + 12+ 22 2 ! . (3)?_,_? 2 + 12 2+ 2 + 12 101 . (4)?_. 4.?A()?“?,y: A()?_?7?k ?Ec, A(c)?_. y: ()?A()?_,K |A()| = a 6= 0 C. ?c C, |A(c)| = a,A(c)?_. 2 ?Cf 181 ()?f() = |A()|,K?c C, f(c) 6= 0,?f()3C ?.f() = a 6= 0 C, |A()| = a 6= 0 C.dA()?_. 5.e?(?: (X,Ky,X?,K?.) ?“?d?7?,k?k K, A(k)?B(k) ?d. ):?.X A() = 1 ! ,B() = 1 2 ! , KA()?B()?d,?k K, A(k)?B(k)?d. 2?Cf 1.e?“?: (1) 2 1 + 12 1 + 12+ 2 + 11 2+ 2+ 3 + 2 2 ; (2) + 112 22 12 12 . ): (1) 3; (2) 2. 2.e?Cf: (1) 110 0 11 00 1 ; (2) + 2( 1)2 + 1 12 0 2 2( 1)22 1 ; (3) + 10 + 01 00 + 00 + ; 182 1?n?“?e?;?/ (4) 1100 0 110 00 11 000 1 ; (5) 0 00 . 0000 ; (6) 000an 100an1 010an2 . 0001 + a1 . ): (1) 1,1,( 1)3. (2) 1, 1,( 1). (3)X 6= 0, 1,1,1,( + )2+ 22;X = 0, 1,1,( + )2,( + )2. (4) 1,1,1,( 1)4. (5)X 6= 0, 1,1, ,1,( )n;X = 0, , , , . (6) 1,1, ,1,n+ a1n1+ + an. 3.?A()?“?,y: rankA() = maxrankA(k)|k K. ):?rankA() = r,KA()k?r?f“Mr+1() = 0.?k ?k K, Mr+1(k) = 0,rankA(k) r.qMr() 6= 0,?3c K Mr(c) 6= 0,r = maxrankA(k) | k K. 4.?Dk() (k = 1,2, ,r)?A()?1?“f,y: D2 k() | Dk1()Dk+1(), k = 2,3, ,r 1. y:?A()?Cf? d1(),d2(), ,dn(), K Dk1() = d1()d2()dk1(), Dk() = d1()d2()dk() = Dk1()dk(), 2 ?Cf 183 Dk+1() = d1()d2()dk+1(), D2 k() = D 2 k1()d 2 k() | Dk1()Dk()dk()dk+1(), D2 k() | Dk1()Dk+1(). 5.?A()?n?,y: A()?AT()?d. y:?3?_?P(),Q(), P()A()Q() = d1() d2() . dn() , ? P()A()Q() = d1() d2() . dn() T = Q()TA()TP()T, uA()?AT()?d. 6. ?f1(x), ,fn(x) Kx,?(f1(x), ,fn(x) = 1. y:?3?“fij(x) Kx (i = 2,3, ,n, j = 1,2, ,n), ? ? ? ? ? ? ? ? ? ? f1(x)f2(x)fn(x) f21(x)f22(x)f2n(x) . . . . . . . . fn1(x)fn2(x)fnn(x) ? ? ? ? ? ? ? ? ? ? = 1. y:?“? A(x) = (f1(x),f2(x), ,fn(x), dfi ?, A(x)?Cf?1,?3?_?P(x), A(x)P(x) = (1,0, ,0).() 184 1?n?“?e?;?/ ?|P(x)| = c 6= 0,K?3?_?Q(x), Q(x)P(x) = 1 1 . 1 c .() P Q(x) = (fij(x), ? B(x) = f1(x)f2(x)fn(x) f21(x)f22(x)f2n(x) . . . . . . . . fn1(x)fn2(x)fnn(x) , Kd(*)?(*)? B(x)P(x) = 1 1 . 1 c , u|B(x)|P(x)| = c,q|P(x)| = c,?|B(x)| = 1,l?fij(x)=?. 3?q? 1.?e?q: (1) A = 325 2610 123 ; B = 62034 63251 42032 . (2) A = 6615 155 122 ; B = 37204 34174 1197011 . 3 ?q? 185 (3) A = 221 111 122 ; B = 133 2613 148 . ): (1); (2); (3). 2.y:?A?=?AT?q. y:duE AT= (E A)T?duE A (SK122.5),dA ?AT?q. 3. ?A?B?n?,y: (AB)= BA y:?“ ?(E + A)(E + B)?(E + A)(E + B)? = |(E + A)(E + B)|E = |E + A|E |E + B|E = (E + A)(E + B)(E + B)(E + A). (E + A)(E + B)(E + A)(E + B) (E + B)(E + A) = 0. ?“?g,? (E + A)(E + B) (E + B)(E + A)= 0, = (E + A)(E + B)= (E + B)(E + A). - = 0?k (AB)= BA. 4. y:XJ?A?B?q,K?A?B?q. y:A?B?q,K?3?_?P,P1AP = B.u|S K4?(?, B= (P 1AP) = P A(P1) = P A(P)1. ?A?B?q. 5. y:?q?*?. y:?A,B?K1?,KE A?E B?Cf? X3K1?“.?K1 K2,o?“?wX 3K2?“,l?Cf|?*?(?f). 186 1?n?“?e?;?/ ?A,B?q?=?E A?E Bk?Cf|.d? ?q?*?. 6. ?A?n?, 0?A?A?.y:A?0? n rank(0E A). y:?0?A?rA?,?d1(), ,dn()?A?Cf. K 0| dn(),? 0 dnr(), (K,X 0| dnr(),K 0| dnr+1(), ,0| dn(),u0? r +1)d?3 ?_?P(),Q() P()(E A)Q() = d1() . dnr() dnr+1() . dn() , ? P(0)(0E A)Q(0) = d1(0) . dnr(0) dnr+1(0) . 0 , dud1(0) 6= 0, ,dnr(0) 6= 0, rank(0E A) rankP(0)(0E A)Q(0) n r. d r n rank(0E A). 4?f 1.e?“?f: (1) 2+ 2 3 12+ 2 3 22+ 3 52 12+ 3 4 2+ 20 1 ; 4 ?f 187 (2) 2 22+ 122 2 2+ 12+ 122 2 2+ 22+ 132 5 . ): (1) , 1, 1, 1, + 3. (2) + 1, 3. 2. fi ?“?A()?f,?r?n,A()?5/: (1) + 1, + 1,( + 1)2, 1,( 1)2; r = 4, n = 5; (2) 2,( 2)2,( 2)3, + 2,( + 2)3; r = 4, n = 4; (3) 1,( 1)2,( 1)3, + 2,( + 2)2; r = 3, n = 5. ): (1) diag(1, + 1,( + 1)( 1),( + 1)2( 1)2,0). (2) diag(1, 2,( 2)2( + 2),( 2)3( + 2)3). (3) diag( 1,( 1)2( + 2),( 1)3( + 2)2,0,0). 3.e?5/: (1) 0( + 1)200 2( 1)000 0002 1 00( + 1)20 ; (2) 2 4000 02+ 200 003 220 0003 4 ; (3) 2+ 2 32+ 200 22+ 2 422+ 300 00 + 1 + 2 002 12+ 2 ; (4) 2 203+ 2 10 2 403+ 22 20 02+ 202+ 6 2 02+ 202+ 5 7 . ): (1) diag(1,( + 1),( + 1)2( 1),2( + 1)2( 1). (2) diag(1,(2 4),(2 4),2(2 4). (3) diag(1, 1,( 1)( + 1),0). (4) diag(1,1,2 4,0). 4.e?Cf,1?“f?f: 188 1?n?“?e?;?/ (1) 425 649 537 ; (2) 213 639 426 ; (3) 1111 1111 . 1111 ; (4) 2300 3400 1512 0223 . ): (1)?Cf: 1,1,2( 1),1?“f: 1,1,2( 1),?f: 2, 1. (2)?Cf: 1,( + 1),1?“f: 1,2( + 1),?f: , + 1. (3)?Cf:1, , |z n 2? ,( n),1?“f:1,2, ,n2, n1( n),?f: , , |z n 1? , n. (4)?Cf: 1,1,1,( + 1)4,1?“f: 1,1,1,( + 1)4,?f: ( + 1)4. 5.?0?n?A?A?,y:?A?uA?0? ?f?un rank(0E A). y:?d1(), ,dn()?A?Cf.XA?uA?0? ?f?r,K 0| dn(), , 0| dnr+1(), 0 dnr(), , 0 d1(). d?3?_?P(),Q() P()(E A)Q() = d1() . dn() , P(0)(0E A)Q(0) = d1(0) . dnr(0) 0 . 0 , 5 e?;?/ 189 u n r = rankP(0)(0E A)Q(0) = rank(0E A), = r = n rank(0E A). 5e?;?/ 1.e?e?;?/: (1) 110 010 121 ; (2) 2615 115 126 ; (3) 131614 676 687 ; (4) 962 18123 1896 ; (5) 133 2613 148 ; (6) 121 242 363 ; (7) 111 333 222 ; (8) 526 203 212 ; (9) 2112 5429 3122 2438 ; (10) 3402 4524 0032 0021 ; (11) 01000 00100 . 00001 10000 ;(12) 123n 012n 1 . 0001 . ): (1) 100 011 001 ; (2) 100 011 001 ; 190 1?n?“?e?;?/ (3) 100 011 001 ; (4) 300 031 003 ; (5) 110 011 001 ; (6) 200 000 000 ; (7) 010 000 000 ; (8) 100 01 + 260 001 26 ; (9) 1000 0110 0011 0001 ; (10) 1100 0100 0011 0001 ; (11) diag(1,1,2, ,n1), 1,1,2, ,n1xn 1?n?; (12) 1100 0110 . . . . . . . . 0011 0001 . 2.? A = 200 a20 bc2 . (1)?A?UkN?e?;?/? (2)(A?z?. ): (1) A=k?A?0= 2,A?e?= A? f?= rank(0E A) (?SK124.5)? rank(0E A) = 2?ac 6= 0, 1?a,c?u0,?u0,a,c?0,?b 6= 0?, 0?a = b = c = 0?. 5 e?;?/ 191 d?ac 6= 0?, A?e?;?/ 210 021 002 ; ?a,c?u0, ?u0,a,c?0,?b 6= 0?, A?e?;?/ 200 021 002 ; ?a = b = c = 0?, A?e?;?/ 200 020 002 . (2) A?z a = b = c = 0. 3.?A?A?“ A() = 5+ 4 53 2+ 8 4. ?Ak?U?e?;?/. ): A() = ( 1)3( + 2)2,dA?U?f?: (a) 1, 1, 1, + 2, + 2; (b) ( 1)2, 1, + 2, + 2; (c) ( 1)3, + 2, + 2; (d) 1, 1, 1,( + 2)2; (e) ( 1)2, 1,( + 2)2; (f) ( 1)3,( + 2)2. ?A?U?e?;?/?: 10000 01000 00100 00020 00002 , 10000 01100 00100 00020 00002 , 11000 01100 00100 00020 00002 , 10000 01000 00100 00021 00002 , 10000 01100 00100 00021 00002 , 11000 01100 00100 00021 00002 . 192 1?n?“?e?;?/ 4. ?A?1.y: A?e?;?/?U? 0 0 . 0 ,X = TrA 6= 0, 01 00 0 . 0 ,XTrA = 0. y:duA?u1,dJA?u1.?A?e?=k ?1,?u0.?0?e?(0), ?1?e?U?()2?e? 01 00 ! .A?e ?;?/?U? 0 0 . 0 , 01 00 0 . 0 . qTrJA= TrA,=?I(?. 5. |?K?(?O?e?1?“: (1) a1xxx xa2xx xxa3x . . . . . . . . . . . xxxan , ai6=x, x6=0; (2) x0a1a2an a0 x1a2an a0a1x2an . . . . . . . . . . . a0a1a2xn , xi6=ai. ): (1) |A|= ? ? ? ? ? ? ? ? a1 x . an x + x 111 . . . . . . . . 111 ? ? ? ? ? ? ? ? 5 e?;?/ 193 = n Q i=1(a i x) ? ? ? ? ? ? ? ? ? ? ? ? ? ? E + x 1 a1 x 1 a1 x 1 a2 x 1 a2 x . . . . . 1 an x 1 an x ? ? ? ? ? ? ? ? ? ? ? ? ? ? = n Q i=1(a i x) ? ? ? ? ? ? ? ? ? ? ? E + x n P i=1 1 ai x 0 0 . 00 ? ? ? ? ? ? ? ? ? ? ? = n Q i=1(a i x) ? 1 + n P i=1 x ai x ? . (2)? |A| = n Y i=0 (xi ai) 1 + n X i=0 ai xi ai # . 6. ?0?n?A?A?.- n0= rankE = n,nk= rank(0E A)k, ak= nk1 nk,bk= ak ak+1, k = 1,2, XeL: n0n1n2n3n4 ? a1 ? a2 ? a3 ? a4 ? b1 ? b2 ? b3 y: (1)?A?uA?0?e?ua1; (2)?A?uA?0?k?e?ubk; y: (1)dSK124.5=?. (2)duni?q?C?,?k?ai,bi?q?C ?.?A?uA?0?k?e?mk,?uA? 0?e?m,K n0= X k1 mkk + m, 194 1?n?“?e?;?/ n1= X k1 mk(k 1) + m, n2= X k2 mk(k 2) + m, nr= X kr mk(k r) + m, l? a1= X k1 mkk + m ! X k1 mk(k 1) + m ! = X k1 mk, a2= X k1 mk(k 1) + m ! X k2 mk(k 2) + m ! = X k2 mk(k 1) + m ! X k2 mk(k 2) + m ! = X k2 mk ar= X kr1 mk(k r) + m ! X kr mk(k r) + m ! = X kr mk(k r) + m ! X kr mk(k r) + m ! = X kr mk br= ar ar+1= X kr mk X kr+1 mk= mr. 7. |?K?(?O?e?e?;?/: (1) 1303 26013 0313 1408 ; (2) 3100 1100 3053 4131 . 5 e?;?/ 195 ): (1)?, 0= 1?A?.?eL: 421000 21100 1010 b1= 1,b2= ,b3= 1. =d?k1?3?e?1?.l?vkO?A? ?.d?e?;?/? 1000 0110 0011 0001 . (2)d?=k1?A?0= 2.?eL: 4200 220 02 ?d?k2?2?e?.d?e?;?/? 2100 0200 0021 0002 . 8. ?A?A?(3E?S)?1.y: Ak?A?q,?, k?(?K?). y:k?A?e?: J = 110 1 . . . 1 01 . 196 1?n?“?e?;?/ ek 0,K Jk= 1k 1 . . . k 01 . uJk?e?= r rank(E Jk) = r(r 1) = 1.Jk? e?;?/?J,l?Jk?J?q. q J1= 11 1 . . . 1 01 . ?n?yJ1?J?q,uJk?Jk?q,l?J?q. u?/,?A?e?;?/? JA= J1 J2 . Js . K Ak Jk A Jk 1 Jk 2 . Jk s JA A. 6?4?“ 1.e?4?“: (1) 123 112 114 ; (2) 252 153 101 ; 6 ?4?“ 197 (3) 3131 1313 3131 1313 ; (4) 1200 2300 0001 0012 ; (5) 111 111 . . . . . . . . 111 ; (6) 123n 1n n12n 2n 1 n 1n1n 3n 2 . . . . . . . . . . . . . . 234n1 . ): (1) ( 2)3. (2) 3 62 4. (3) 2. (4) ( + 1)2. (5) ( n). (6)? P = 0100 0010 . . . . . . . . . . . 0001 1000 ,f() = 1 + 2 + + nn1. K A = E + 2P + 3P 2 + + nP n1 = f(P). duP?A?1,2, ,n1,?ng?.A?A ?“? A() = ( f(1)( f()( f(n1) = ? n(n + 1) 2 ? g(). ? f() = nn+1 (n + 1)n+ 1 (1 )2 , 198 1?n?“?e?;?/ f(k) = n n (1 k)2 = n 1 k .() g() = n1 Y k=1 ? + n 1 k ? = 1 n1 Q k=1(1 k) n1 Y k=1 ( + n k) = 1 n n1 Y k=1 ? + n k ? = n1 n ? + n ?n 1 + n 1 = n1 n2 ? + n ?n 1 ? = 1 n2 ( + n)n n. A() = 1 n2 ? n(n + 1) 2 ? ( + n)n n. qd()?, A()?,?A?4?“?A?“,l?A? 4?“? 1 n2 ? n(n + 1) 2 ? ( + n)n n. 2.?A?n?, m()?4?“, g()?“, d() = (m(),g(). y: (1) rankd(A) = rankg(A); (2) g(A)?_?7?g()?m()p?; (3)Xg(A)?_,Kg1(A)?A?“. y: (1)?3?“u(),v(), m()u() + g()v() = d(). ? m(A)u(A) + g(A)v(A) = d(A). dm(A) = 0?d(A) = g(A)v(A). rankd(A) rankg(A). qd() | g(),?3h()d()h() = g(),=d(A)h(A) = g(A).u rankg(A) rankd(A). 6 ?4?“ 199 ? rankg(A) = rankd(A). (2) ()?1, ,n?A?A?.Kg(A)?A? g(1), ,g(n).Xg(A)?_,Kg(A)?z?A?g(i) 6= 0.dum() ?A?A?,dg()?m()?,l?(g(),m() = 1. ()X(g(),m() = 1,Kd(