线性代数习题答案(复旦版周勇_朱砾)

线性代数习题及答案all in

习题一

1. 求下列各排列的逆序数.

(1) 341782659； (2) 987654321；

(3) n(n?1)…321； (4) 13…(2n?1)(2n)(2n?2)…2. 【解】

(1) τ(341782659)=11; (2) τ(987654321)=36;

n(n?1)(3) τ(n(n?1)…32221)= 0+1+2 +…+(n?1)=;

2(4) τ(13…(2n?1)(2n)(2n?2)…2)=0+1+…+(n?1)+(n?1)+(n?2)+…+1+0=n(n?1).

2. 略.见教材习题参考答案. 3. 略.见教材习题参考答案.

5x124. 本行列式D4?3x1xi1i2i3i4x12的展开式中包含x3和x4的项.

2x3122x(i1i2i3i4)解： 设 D4??(?1)?ai11ai22ai33ai44 ，其中i1,i2,i3,i4分别为不同列中对应

元素的行下标，则D4展开式中含x3项有

(?1)?(2134)?x?1?x?2x?(?1)?(4231)?x?x?x?3??2x3?(?3x3)??5x3

D4展开式中含x4项有

(?1)?(1234)?2x?x?x?2x?10x4.

5. 用定义计算下列各行列式.

0200123000100020(1)； (2). 3000304500040001【解】(1) D=(?1)τ(2314)4!=24; (2) D=12.

6. 计算下列各行列式.

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214?1?ac?ae(1)

3?12?1ab123?2； (2) ?bdcd?de； 506?2?bf?cf?efa?1001234(3)1b?102301c?1； (4) 413412. 001d4123506?2【解】(1) Dr1?r23?12?1123?2?0; 506?21?1?1(2) D?abcdef?11?1??4abcdef;

?1?1?1b?101?10(3)D?a1c?1?(?1)20c?1?a??bc?1?1?1???cd?101d01d?1d0d?

?abcd?ab?ad?cd?1;102341023410234c1?c(4)D210341r2?r1011?3rc??3?2r2011?3c1?cc310412rr3??rr102?2?2r?4?r200?44?160.1?410123410?1?1?1000?47. 证明下列各式.

a2abb2(1) 2aa?b2b?(a?b)2；

111a2(a?1)2(a?2)2(a?3)2(2)

b2(b?1)2(b?2)2(b?3)2c2(c?1)2(c?2)2(c?3)2?0；

d2(d?1)2(d?2)2(d?3)21a2a31aa2 (3) 1b2b3?(ab?bc?ca)1bb2 1c2c31cc2 2

a00b(4) D0ab02n?0cd0?(ad?bc)n；

c00d1?a111(5)

11?a21??n?1?n?1?????ai. i?1aii?1111?an【证明】(1)

c?c(a?b)(a?b)b(a?b)b2左端1c?3?c2(a?b)a?b2b23001

?(a?b)(a?b)b(a?b)a2(a?b)a?b?(a?b)2?bb21?(a?b)3?右端.a22a?14a?46a?9a22a?126c2-c1b2(2) 左端2b?14b?46b?9c3-2cb22b?126c?c3?2?cc12c?14c?46c?9c?24?3c241cc22c?126?0?右端.d22d?14d?46d?9d22d?126(3) 首先考虑4阶范德蒙行列式:

1xx2x3f(x)?1aa2a31bb2b3?(x?a)(x?b)(x?c)(a?b)(a?c)(b?c)(*)1cc2c3从上面的4阶范德蒙行列式知，多项式f(x)的x的系数为

1aa2(ab?bc?ac)(a?b)(a?c)(b?c)?(ab?bc?ac)1bb2,

1cc2但对（*）式右端行列式按第一行展开知x的系数为两者应相等，故

1a2a3(?1)1?11b2b3, 1c2c3(4) 对D2n按第一行展开，得

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aabD2n?ac0据此递推下去，可得

b00aab?bcd0cc0bcdd0

d00d?ad?D2(n?1)?bc?D2(n?1)?(ad?bc)D2(n?1),D2n?(ad?bc)D2(n?1)?(ad?bc)2D2(n?2)??(ad?bc)n?1D2?(ad?bc)n?1(ad?bc) ?(ad?bc)n?D2n?(ad?bc)n.

(5) 对行列式的阶数n用数学归纳法.

当n=2时，可直接验算结论成立，假定对这样的n?1阶行列式结论成立，进而证明阶数为n时结论也成立.

按Dn的最后一列，把Dn拆成两个n阶行列式相加：

Dn?1?a1111?a211an?1?anDn?1.1111111?a11?1111?a211111?an?11000an

?a1a2但由归纳假设

Dn?1?a1a2从而有

Dn?a1a2?a1a2?n?11?an?1?1???,

?i?1ai??n?11?an?1?ana1a2an?1?1????i?1ai? nnn1??1??an?1an?1?????1????ai.?i?1ai??i?1ai?i?18. 计算下列n阶行列式.

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(1) Dn?x11x11x0yx000y0011x1222222232222； n (2) Dn?2200x000(3)Dn?0yyx. (4)Dn?aij其中aij?i?j(i,j?1,2, ,n) ；

2101210120000000000002112(5)Dn?.

【解】(1) 各行都加到第一行，再从第一行提出x+(n?1),得

11Dn?[x?(n?1)]1x11将第一行乘（?1）后分别加到其余各行，得

11x,

Dn?[x?(n?1)]111x?1002000n?210x?1?(x?n?1)(x?1)n?1.

12221000r2?r1(2)

Dn?101010021000r3?r1 rn?r1按第二行展开222010?002000

200n?25

??2(n?2)!.

T(?2，0，1，0，0),(?1，?1，0，1，0).

(4) 方程的系数矩阵为

??12?22?1??12?22?1?A??12?13?2?????r2?r1r3?2r1??0011?1???24?711??????00?3?33???12?22?1?

???r3?3r2???0011?1?R(A)?2,??00000???∴ 基础解系所含解向量为n?R(A)=5?2=3个

?x2??取??x??x2?4为自由未知量 ?????0??1??0?x4??0?,?0??x5???????x5??????,??1??, ?1???0????0????3???2????4?0得基础解系 ?0?????1???1?,?0,????0?????0?????1?1? .???1????0?????0??3. 解下列非齐次线性方程组.

??x1?x2?2x3?1,(1) ??2x1?x2?2x3?4,?2x1?x2?x3?x4?1, (2) ??4x1?2x2?2x3?x4?2,?x1?2x2?3,???4x?4x?2x1?x2?x3?x4?1;1?x23?2;?x?2x?x?x?1,?x1?x2?x3?x4?x5?7,(3) ?1234??x?x?3x1?2x2?x3?x4?3x5??2,?1?2x23?x4??1, (4) ??x1?2x?x3?xx?2x3?2x4?6x5?23,4?5;?22??5x1?4x2?3x3?3x4?x5?12.【解】

（1） 方程组的增广矩阵为

??1121??1121?(Ab)??2?124??r2?2r0?3?22???1?203?????1r3?r1???22?????r3?r2r4?r2??4142?r4?4r1?0?3???0?3?4?2????1121??121?

?0?3?22??1r?14?r30?3?22???0000??????2???0012???00?2?4????0000??

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得同解方程组

x?2,???x1?x2?2x3?1?3??3x?2x?2?2?2x3?23??x2??x?3??2, 3?2???x1?1?x2?2x3??1.(2) 方程组的增广矩阵为

?21?111??21?111?(Ab)??r3?r?42?212?????1?r2?2r1?000?10? ??11????21?1???000?20??得同解方程组

??2x1?x2?x3?x4?1,??x?4?0,?x4?0 ??2x4?0,即

??2x1?x2?x3?1,?x 4?0.令x1?x3?0得非齐次线性方程组的特解

xT=(0，1，0，0)T.

又分别取

??x2??1?????0?x???0??,??1?? 3得其导出组的基础解系为

T?1?T1?????2,1,0,0??;??1?2???2,0,1,0??,

∴ 方程组的解为

??0????1??1?2??x??1????k???2??k1,k2?R

?0?1?1???k2?0?.?0?0??1?????0??????0????1?2111??1?2(3) ??1?21?1?1?111????r2?r1??000?2?2?1?2115?r3?r1????????00004??R(A)?R(A)∴ 方程组无解.

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(4) 方程组的增广矩阵为

?1?3(Ab)???0??5?1?0r3?r2?????r4?r2?0??0分别令

11117?7??11111?0?1?2?2?6?23?211?3?2?r3?3r1???????r4?5r1?0122623?122623????433?112?0?1?2?2?6?23??

11117??1?2?2?6?23??,00000??00000??x3??0??1??0??x???0?,?0?,?1? ?4?????????1????0????0???x5????x?x?x?x?x?0得其导出组?12345的解为

??x2?2x3?2x4?6x5?0?5??1??1???6???2???2???????k1?0??k2?1??k3?0???????00?????1?????1???0???0??k1,k2,k3?R.

令x3?x4?x5?0,

得非齐次线性方程组的特解为：xT=(?16,23,0,0,0)T,

∴ 方程组的解为

??16??5??1??1??23???6???2???2?????????x??0??k1?0??k2?1??k3?0? ????????000???????1??????0???1???0???0??其中k1,k2,k3为任意常数.

4. 某工厂有三个车间，各车间相互提供产品（或劳务），今年各车间出厂产量及

对其它车间的消耗如下表所示. 车间 出厂产量 总产量 1 2 3 消耗系数 （万元） （万元） 车间 1 0.1 0.2 0.45 22 x1

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2 0.2 0.2 0.3 0 x2 3 0.5 0 0.12 55.6 x3 表中第一列消耗系数0.1,0.2,0.5表示第一车间生产1万元的产品需分别消耗第一，二，三车间0.1万元，0.2万元，0.5万元的产品；第二列，第三列类同，求今年各车间的总产量.

解：根据表中数据列方程组有

??x1? 0.1x1?0.2x2? 0.45x3?22,?x2? 0.2x1?0.2x2?0.3x3?0, ??x3?0.5x1?0.12x3?55.6,?0.9x1?0.2x2? 0.45x即 ?3?22,? 0.2x?1?0.8x2?0.3x3?0,

?0.5x1?0.88x3??55.6,?x1?100解之 ?,?x?2?70,

?x3?120;5. ?取何值时，方程组

???x1?x2?x3?1,?x1??x2?x3??, ??x1?x2??x3??2,(1)有惟一解，(2)无解，(3)有无穷多解，并求解.

【解】方程组的系数矩阵和增广矩阵为

???11??A??1?1??;??111?B???1?1??,

?11?????11??2???|A|=(??1)2(??2).

(1) 当?≠1且?≠?2时，|A|≠0，R(A)=R(B)=3.

∴ 方程组有惟一解

???11(??1)2x1???2,x2???2,x3?(??2).

（2） 当?=?2时，

???2111??1?21?2?B??1?21?2??24????r2?r1??r???2111?????3?r1r2?2r1???11???11?24????1?21?2??

?0?33?3??1?21?2??0??03?36????33?3??,???0003??

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R(A)≠R(B),∴ 方程组无解. (3) 当?=1时

?1111??1111?r2?r1?0000? B??1111???????r3?r1?????1111???0000??R(A)=R(B)<3,方程组有无穷解.

得同解方程组

?x??x?x?1,?123?x2?x2, ??x3?x3.∴ 得通解为

??x1??????1???1??1?x2?k1??k?0???0?, k1,k?R.

??1??x3???0?2?????2????1????0??6. 齐次方程组

???x?y?z?0,?x??y?z?0, ??2x?y?z?0当?取何值时，才可能有非零解?并求解. 【解】方程组的系数矩阵为

??11?A???1??1???

?2?11??|A|=(??4)(??1)

当|A|=0即?=4或?=?1时，方程组有非零解.

(i) 当?=4时,

?411??14?1??14?1?A???14?1????r2?r1??41????1????r2?4r1r3?2r1??0?155??2?11????2?11?????0?93????14?1??14?1????r12?5?r10?31?r3??31?3?3???0?31????r2????0???000???得同解方程组

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