a11000000a11a22a23a24a12a22a32a42a22a32a42a12a32a33a34a13a23a33a43a23a33a43a23a42a14a24a34a44a24a34a44a24?a11?(?1)4?1B??a11B??1
?a11?(?1)1?1B?1?????????a11?0,且B?1?0 a111a43?BT?B?.
a11a44123（15）设有行列式?10式A12=??????1?1?223x0，则元素?1的余子式M21=????????，元素2的代数余子
x1x1?10????； 012341341241=?(aij)，Aij表示元素aij的代数余子式，则 2312（16）设D?34A14?2A24?3A34?4A44? 0 ；
解：方法一：A14?2A24?3A34?4A44可看成D中第一列各元素与第四列对应元素代数余子式乘积之和，故其值为0.
12方法二：A14?2A24?3A34?4A44?34ac（17）设D?dabbbbcdcd234134121234推论10.
da=?(aij)，Aij表示元素aij的代数余子式，则 acA14?A24?A34?A44? 0 ；
5
ac解：A14?A24?A34?A44?da54（18）设f(x)?bc1bd1推论40.
bc1bd120000x0000000064320020003x00x00000002?x00，则x5的系数为 6 ；
3x02?x0解：方法一：
54f(x)?3x0方法二：
43203x02?x0054?63x43203x02000x0?6?(?1)005?422?x02?x0?(?1)2?x5?6x5
2?x000006f(x)只有一项非0
54?f(x)?3x0432003x0020000x000000006???1?t?543216?2?x00a15a24a33a42a51a66
2?x0?????????????(?1)10?(?1)2?x5?6?6x5综上所述：x5的系数为6.
0（19）设D?a11a21a12a22am1am2a11a12a22am2a1ma2mamm?a
am1am2b11b12b1nb2nbnnc11c21c12c22b21b22bn1bn2ammc1mc2mcnm， 且
a21am1cn1cn26
b11b12b1nb2nbnna11a12a22am2a1ma2mamm?a，B?b11b12b1nb2nbnn?b
a21am1b21b22bn1bn2?b，则D=?????????1?ab???????；
mnb21b22bn1bn2解：方法一：令A?则D1?AOOAmnmn?A?B?ab,D2????1?A?B???1?ab
CBBC证明：根据行列式性质2和5，将行列式
A变成下三角行列式，得到：
a11A?a21am1a12a22am2a1ma2mamm?a1a21?am1?a2am2?am?a1a2am?a
行列式D1、D2的变换和行列式
A的变换完全相同，得到：
a1a21?am1?c11?c21?cn1?a2am2?c12?c22?cn2?amc1m?b11b12b1nb2nbnnc2m?b21b22cnm?bn1bn2D1?
7
a1a21?am1?b11b21b12b22b1nb2nbnnc11?c21?cn1?a2am2?c12?c22?cn2?amc1m?c2m?cnm?D2?
bn1bn2分别将D1、D2第一次按第一行展开（a2变成第一行），第二次按第二行展开（a3变成第一行），……，总共进行m次第一行展开，得到：
D1?a1a2amB?A?B?ab；
1?n?1D2?a1??1?证毕.
?a2??1?1?n?1am??1?1?n?1?B???1??A?B???1?ab
mnmn?AO?方法二：设A??aij?，B??bpq?，D?????dij??m?n???m?n? m?mn?nCB???aij, i?1:m,j?1:m ?其中：dij??bpq, i?m?1:m?n,j?m?1:m?n,p?i?m,q?j?m ???
?c, i?m?1:m?n,j?1:m,p?i?m ?pj那么：D???由?????AOCB??p1,tp1,pm?n???1,,m?n?pm?m?l1????1???m?ln??tp1pmpm?1pm?n?d1p1dmpmdm?1,pm?1dm?n,pm?n
?p1,?l1,,pm?1,,m?,lm???1,,n??????1??a1p1ampmb1l1t?l1ln?bnlnbnln????p1,?l1,,pm?1,??????1?t?p1?,m?pn?a1p1ampm???1?b1l1,lm???1,,n?pn?
?t?p???1?1????p,,p???1,,m??1m?A?B?aba1p1??ampm??????l,??1,lm???1,,n????1?t?l1ln?b1l1?bnln???8
a11a12a1ma21a22a2mD?am1am2amm2b11b12b1nc11c12c
1mb21b22b2nc21c22c2mbn1bn2bnncn1cn2cnmD2中am?依次与b1?,b2?,,bn?对换，使得am?在bn?下面；
a?m?1??依次与b1?,b2?,,bn?对换，使得a?m?1??在bn?下面，在am?上面；……
a1?依次与b1?,b2?,,bn?对换，使得a1?在bn?下面，在a??上面；
总共进行了mn次对换。得到：
DC2???1?mnBOA???1?mn?B?A???1?mnab.
1?aa000?11?aa00（20）D5=0?11?aa0=????1?a?a2?a3?a4?a5????.
00?11?aa000?11?a1?aa0001a00?11?aa0001?aa0解：D5?0?11?aa0c1?c2?c3?c4?c50?11?aa00?11?aa00?11?a000?11?a?a00?1?????按第一列展开D4?a?(?1)5?1?a4?D4?a5
同理可得：D4?D3?a4，D3?D2?a3，D2?D1?a2 则D5?D1?a2?a3?a4?a5?1?a?a2?a3?a4?a5. 2．选择题
9
000a?a
1