a2b2c22d3)
(a?1)2(b?1)2(c?1)2(d?1)2(a?2)2(b?2)2(c?2)2(d?2)2(a?3)2(b?3)2(c?3)2(d?3)2
解: 下面都将所求行列式的值设为D.
1) 因为第1行加到第2行以后, 第2行将和第4行相等, 因此行列式的值D=0; 2) 首先从第1,2,3行分别提取公因子a,d,f, 再从第1,2,3列提取公因子b,c,e, 得
?abbdbfac?cdcfae?111?11102?4abcdef20
de?adfbce1?ef1?11?abcdef01?103) 将第2,3,4列都展开, 并统统减去第1列, 得
a2D?b2c2d2a2b2D?2cd22a?14a?46a?92b?14b?42c?14c?46b?96c?9再将第3列减去2倍的第2列, 第4列减去3倍的第2列, 得
2d?14d?46d?9 2a?1262b?126?02c?1262d?126
#. 把下列行列式化为上三角形行列式, 并计算其值
?243 2解:
2?40?1351?2?3051;
r1?2?r2?40r1?(3/2)?r3?22?40?235r1?1?r403?55c2?c30????2?304?8?305102110?420r2?8?r3?2?420?2121r2?5?r40121c3?c40????84?3002050?5350013100?2?402r3?(?2)?r40112???2?1?5?(?27)??27000520000?27 ?224?13120?2r2?r40?0016
?4?5?81?410020354?321 02125201013
#. 计算下列n阶行列式
123234345???n12abab???ba
?????1) n12?n?1 1?2?3???n?2) b解: 1) 设此行列式的值为D, 将第2,3,…,n列均加于第一列, 则第一列的所有元素均为
1n(n?1)2, 将此公因式提出, 因此有
123?n再令第n行减去第n-1行, 第n-1行减去第n-2行, …, 第2行减去第1行, 可得
134?11D?n(n?1)145?22?????112?n?1
101D?n(n?1)02?01n?1?????????????23?n11?11?n11?1?n11?1?n1111?1?n(n?1)?????2????11?n?111?n1?11?n1?11 1?10?0?n11?10??n1?n(n?1)(????????211?1?n?02) 此题和第3题的2)一样, 因此有D?a?(?1)
#. 证明下列行列式
n1n(n?1)012)?n(n?1)(?1)(?n)n?1?20
n?1bn
2n?????????????aba?abba?b?a
b??(a2?b2)na2) b证:
用归纳法, 设Dn为所求行列式值, 当n=1时,
17
D1?ab?a2?b2ba, 等式成立. 2k?????????????aba??abba?bb?aa
假设当n=k时假设成立, 即有
bDk??(a2?b2)k当n=k+1时,
2k??2????????????aba?Dk?1??bb2?k??1?????????ab?ab?a?b22b?abba?aa2?k??1?????????ba??b?aa22按第一列展开?
?bab?abba??a22k222?b
证毕.
?aDk?bDk?Dk(a?b)?(a?b)(a?b)?(a?b2)k?1
3??20?A??1?11????01?2??的伴随矩阵A*, 并求A-1. #. 求矩阵
解:
A11?18
?11?1,1?2A21??03?3,1?2A31?03?3?11
11?2,0?21?1A13??1,01A12??A22?23??4,0?220A23????2,01A32??23?111 20A33???21?1
3??A11A21A31??13???2?41?A*??AAA122232??????A13A23A33????1?2?2?? 因此得
A的行列式为|A|?a11A11?a12A12?a13A13?2?1?0?2?3?1?5 3??131*1?A?1?A??2?41??|A|5??1?2?2?? 因此有
#. 设A为n阶可逆阵, A2=|A|I, 证明: A的伴随矩阵A*=A. 证: 因A可逆, 则在等式A2=|A|I两边乘A-1, 得A=|A|A-1, 即
A?1?11*AA?1?A|A|, 而因为|A|, 所以有A=A*, 证毕.
#. 用克莱姆法则解下列方程组.
解:
方程的系数矩阵A为
?2x1?3x2?11x3?5x4?6?x?x?5x?2x?2?1234??2x1?x2?3x3?4x4?2??x1?x2?3x1?4x2?2
?2?1A???2??13115?152??134??134?, 常数向量???6222?T
先求A的逆A-1:
?2?1??2??131151152013401340r1?(?2)?r2?1r1?(?2)?r3?r1?(?1)?r4?0???????0??001000??115?23110?r?r12????????2130???1??1131520100?1111?200???1?700?210??0?220?101?
001025440100100000100?0??0??1?
19
?1r2?(?1)?r1?0r2?1?r3????????0??0?1r3?r4?r3?(?1/2)?0???????0??0r3?(?4)?r1?1r3?(?1)?r2?0r3?6?r3????????0??0?1?r4?(?1/5)?0???????0??0r4?(?5)?r1?1r4?(?2)?r2?0r4?1?r3????????0??00411110?610?2204111101?10?6100100152?1?1110?1101?13?2?4?13?21/2?41001000010?0??0??1? 0?0???1/2??0?
000?5010005021?1010001000100010?0?7/5?29/10A?1????1/57/10?1/5??1/5因此有
01?1??6??0??x1??0?x??7/5?29/102/5?7/10??2??2?2?1??????X????A????x3???1/57/10?1/51/10??2??0?????????x?1/51/5?1/53/5???2??0? 4??则
即x1=0, x2=2, x3=0, x4=0.
#. 如果齐次线性方程组有非零解, k应取什么值?
2?1?5/201/2??01/20?1/2??1?11?3? ?1102?1?5/201/2??01/20?1/2???1/51/5?1/53/5? 001?1?7/5?29/102/5?7/10???1/57/10?1/51/10???1/51/5?1/53/5? 1?1?2/5?7/10???1/51/10???1/53/5?
解: 此方程组的系数矩阵A为
?(5?k)x?2y?2z?0??2x?(6?k)y?0?2x?(4?k)z?0? ?5?kA???2??226?k0要使方程组有非零解, 必须有det(A)=0.
20
2?0??4?k??