线性代数机械工业出版社第一章答案

线性代数第一章行列式

一、填空题 1.排列631254的逆序数?(631254)= 8 . 解: ?(631254)=5+2+1=8

1232.行列式231= -18 . 312解:D=1?3?2+2×1×3+2×1×3-3?3?3-1?1?1-2?2?2=-18 3、4阶行列式中含a12a24且带正号的项为_______ 答案：a12a24a33a41

分析：4阶行列式中含a12a24的项有a12a24a33a41和a12a24a31a43 而 a12a24a33a41的系数：??1??(1234)??(2431)?(?1)4?1 ?(?1)3??1

a12a24a3a??1?1的系数：4?(1234)??(2413)因此，符合条件的项是a12a24a33a41

1aa224、1bb（a,b,c互不相等）=_______

1cc2答案：(b?a)(c?a)(c?b)

1aa22222222分析：1bb=bc?ab?ac?bc?ac?ba?(b?a)(c?a)(c?b)

1cc2

?15.行列式

07454001?1004?370?（?1）=(－1) ×7×6×(－1)=42 46123106中元素?的代数余子式的值为 42

?1?解析: 元素?的代数余子式的值为1226.设D3?12230,则代数余子式之和A21?A22?A23=0 2-132解析：A21?A22?A23=1×A21+1×A22+1×A23=1二、 单项选择题

2211=0 2?132x11、设f(x)?31xx211231?1，则x的系数为（C）

x11x?（1234）??（2134）A. 1 B. 0 C. -1 D. 2 解：

x3的系数为

（-1）=-1

a112、 设a21a31a11解：a21a31a12a22a32a12a22a32a133a11a23=m?0,则3a21a333a31a13a11a23 j2?(?4)?a21a33a312a11?4a122a21?4a222a31?4a32-4a12-4a22-4a32a13a23=（B） a33A.12m B. -12m C.24m D. -24m

a13a23 =-4m a33a112a11-4a122a21-4a222a31-4a32a13a13a23 =-4m a33j1?2?j2?a21a313a112a11-4a122a21-4a222a31-4a32j1?3?3a213a313.行列式

a23 =-12m a33k-12?0的充分必要条件是（C）

2k-1(A.)k?-1 (B)k?3

(C)k?-1且k?3(D)k?-1或k?3 因为原式=(k-1)(k-1)-4?0 所以k-1?2且k-1?-2 所以k?-1且k?3 所以答案为C

a04.行列式

0df0b00h中元素g的代数余子式的值为（B）

0c0eg0（A）bcf-bde (B)bde-bcf (C)acf-ade (D)ade-acf

00dfb0=-(bcf-bde)=bde-bcf 0）A(?1=414+1ce所以答案为B

a115.设D=

a12a22...an2n...a1na21...an1?ka11...a2n?ka21,则

............ann?kan1n?ka12?ka22...?kan2n...?ka1n...?ka2n=( )

.........?kann(A)-kD (B)-kD (C)kD (D)(-k)D 答案:D

?ka11解:由行列式性质3:将

?ka12?ka22...?kan2...?ka1n...?ka2nn的每行提出一个-k,得到(-k)D,即为选项D.

.........?kann?ka21...?kan16.行列式D1000...=09000...800..................02...0001000......=( )

0000010(A)50 (B)-(10!) (C)10! (D)9! 答案:C

解:由行列式的定义,每个因式的元素取自不同行不同列,且不为零,则每行依次取出1,2,…,10,得到10!.又因为

?(98765432110)?36为偶数,所以结果为正数.最终结果为10!

三、计算题

11、计算行列式D?202310420?54?2?12?9?2?2?2?7?10?120?3?9111=?6035013.

113?1?11230?2?2=

0?7?1000?3111112001030100412341012解D=

3?1?10320?5=1*?1???1?1?=?24

2、计算行列式D?.

11111200解、D=

11111?1?101?1?1?1?1??12?1=?2 ==1*??1?10300?12?110040?1?13?1?1311143.计算行列式D?11311211

111111140003解 D?11310021211=00100= -6 1111111112344.计算行列式D?23413412

412312341023410234解 D?234111?33412cc1034101?i(i?2,3,4)10412ri?r1(i?2,3,4)00?44 =160 412310123000?45. 计算n阶行列式

x?(n?1)aaa...a1aa...ac1?cix?(n?1)axa...axDin(i?2,?3,...,n)...............=[x+(n-1)a]

1xa...ax1?(?1)?...............(i?2,?3,...,n)x?(n?1)aaa...x1aa...x1aa...a0x?a000...............=[x+(n-1)a] (x?a)n?1

000...x?a?3x1?2x2?x3?6.当k为何值时，方程组?0?kx1?7x2?2x3?0有非零解.

??2x1?x2?x3?0解由题知

32-1r32?11?(?2)?r2D=k7-22-13r1??3?rk?630=(?1)?(?1)1?3k?63=-5(k-6)+33=0 得k=63 311501155四.解答题

[x+(n-1)a]

2?1012中第3列元素的余子式和代数余子式的值，并求出D的值。 1.写出D=4?11?1解：M?1012 A3?131==-2 31=(-1) ×(-2)=-2

M20?232=

42=4 A32=(-1)

3×4=-4

M2?133=

41=6 A3?3

33=(-1)×6=6

D=-1×(-2)+1×(-4)+(-1)×4=-8

?2、用Cramer法则解线性方程组?2x1?x2?3x3?1?2x2?x3?5

??6x1?x2?x3?3213213解D=02?1=02?1=-40

6?110?4?8113213211且D1=52?1=-40 D2=05?1=-80 D3=025=40 3?116316?13所以x1=1 x2=2 x3=?1 五、证明题

a1?b1ia1i?b1c1a1b1c1 1.设i2??1，试证：a2?b2ia2i?b2c2?2a2b2c2

a3?b3ia3i?b3c3a3b3c3a1?b1ia1?i1b1ca1?a1i1b1c1bi?1ai1证：a2?b2ia2?ib2=c2a2?a2i2b?2c2bi?2ai2

a3?b3ia3?ib3c3a3?a3i3b3c3bi?3a3

b

a1a1ic1a1b1c1b1ia1ic1b1ib1c1?a2a2ic2?a2b2c2?b2ia2ic2?b2ib2ca12a3a3ic3a3b3c3b3ia3ic3b3ib3c?0?a23a3又因为i2??1,

1c2c3cb1c1a1b2c22?ia2b3c3a3b1c1b2c2?0b3c3bb a1b1c1所以原式=2a2b2c2， a3b3c3所以证毕

??x1x2??xn?1?axa2??anxn?b 2.设a?a?11x2221,a2,n互不相同，证明：线性方程组?a1x1a??a2?2x2nxn?b2 ????????an?11xn?11a2x2??an?1?bn?1nxn11?1a1a2?an 证：系数行列式为范德蒙行列式D=a21a222?an?j)

????1??(ai?ai?j?nan?1n?11a2?an?1n 因为a1,a2,?,an互不相同，

所以D?0，

故该线性方程组有唯一解，

证毕

a11a12a133

设

a21a22a23 =a,

b11b12ab=b,证21b2231a32a332a11a11?a12a12a11?a122a11a11a11a12a11a122aa22a?a1221a21?a??2221?a222a21a21a21a?a1222a22a21a22a=72ab.

22002a112a122a1300a11a12a13002a212a222a2300a21a22a23解:

002a312a322a33=7200a31a32a33 3b113b12000b11b120003b213b22000b21b22000由拉普拉斯展开定理可知

00a11a12a13002a112a1200a21a22a23a11a12a13000a31a32a33=a21a22ab2a212a2211b01223?=ab 所以002a312a32b11b12000a31a32ab21b22333b113b1200b21b220003b213b2200

明2a132a232a33=72ab 00：

本文来源：https://www.bwwdw.com/article/e6yt.html