x1x1?x21111?x2x
3x411114?x11x2x3x4?k?1x. k?14.设A??171??2?12?,则A5?43??A? . ?020??
三、应用举例
例2 用初等变换法计算行列式
2?11?31?73?39?5D?402?21 7?53?14610?44?102解：
1?12?31cc?33?79?5D1??3?204?213?57?1464?410?102 11
r1?12?31r4?r2r2?32rr1r3?100?10?2???4?4r1?0204?1
0?20?510022?21?12?311?12?31rr2?r30204?10?21?53???4?r200?10?2r5?2r???3001?12
000?10000?100022?20002?61?12?31r0?21?53???5?2r4001?12000?10???2???1???6???12.
0000?63?521例3 设 D?110?5?1313，求
2?4?1?32A11?5A21?3A31?3A41 及 3M31?M32?M33?2M34的值 .
2111解 2A?3A510?511?5A21?3A3141??3313
3?4?1?3
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3111c1?r?4010?50313
0?4?1?3按C10?51展开r10?5?3?(?1)1?13133?r?23313?4?1?3?100按C2展开?3?(?1)2?21?5?10??15.
3M31?M32?M33?2M34
?3A31?A32?A33?2A343?5219?705?110?5rr1?2r34?r33?1?12?110?53?1?12 2?4?1?3?1?30?5按c5r3展开9?7r1?r38?100?（?1）（??1）3+311?52?r?3?240?1?3?5?1?3?5按c3展开?-（?5）（??1）3+38?1024?5(32?20)?260.
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2111510?5提问14：若D?，如何计算5A13?2A23?2A43?33133?4?1?3值.
abb?bbab?b例4 计算行列式 D?bba?b
?????bbb?a解 此为循环行列式，将第2,3,?,n列都加到第一列得
ca??n?1?bbb?bc1?c21?c?3c?a?1?cn?n?1?bab?bD?a??n?1?bba?b
?????a??n?1?bbb?a提取公因式得
1bb?b1ab?bc1??a?(n?1)b??[a??n?1?b]1ba?b
?????1bb?a 14
b1b?br2?rr1??3?r1rn?r0a?b0?0?1?a?(n?1)b?00a?b?0
?????000?a?b??a?(n?1)b?(a?b)n?1.（提问：是否还有其他方法？）
例5 计算行列式
x?a1a2a3?ana1x?a2a3?anDn?a1a2x?a3?an(x?0)
?????a1a2a3?x?an解 加行加列得
1a1a2?an Dn?Dn?1?0?D
n01a1a2?anr(i?2i?r?1?1x0?0?,n)?10x?0 ??????100?x【将上述箭形行列式化为三角形行列式】
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