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)…3·2·1)= 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.
5x1232的展开式中包含x3和x4的项. 3x1xx12x4. 本行列式D4?122x(i1i2i3i4)解： 设 D4?i1i2i3i4?(?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.
1
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?01d01d?1d0d??cd?1? ?abcd?ab?ad?cd?1;102341023410234cr(4)D1?c?2103412?r1011?3r3?2r11?3c10412r?2?2?2r?204?160.c1?c3r3?4?r200?41?c4?rr1010123410?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) D002n?ab0cd0?(ad?bc)n； c00d1?a111(5)
11?a21?n?1?n??1??a?i??ai. i?1i?1111?an【证明】(1)
c(a?b)(a?b)b(a?b)b2左端1?c3c?2(a?b)a?b2b2?c3001
?(a?b)(a?b)b(a?b)2(a?b)a?b?(a?b)2a?bb21?(a?b)3?右端.a22a?14a?46a?9a22a?126(2) c2-c左端1b22b?14b?46b?9c3-2c2b22b?126c?c3?c124?c1c2c?14c?46c?9c?4?3c2c22c?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
3
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：
1?a1111?a21?a1a21an?1?anDn?1.1111111?a11?11111?an?110an11?a21100Dn?
但由归纳假设
Dn?1?a1a2从而有
?n?11an?1?1???i?1ai??, ?Dn?a1a2?a1a28. 计算下列n阶行列式.
?n?11?an?1?ana1a2an?1?1????i?1ai?
nnn1??1??an?1an?1?????1????ai.?i?1ai??i?1ai?i?1 4