高等代数作业 第二章行列式答案 下载本文

n?a1nDn?1?a1a2?an?1?a1a2?an(1??). i?1ai112312?x25. 解方程:232315=0.

2319?x2112311231123解:12?x22301?x200002315=

01?3?1=(1?x2)?0101?3?1

2319?x201?33?x201?33?x211231123=(1?x2)?0100000?3?1=(1?x2)?10000?3?1=?3(1?x2)(4?x2)

00?33?x20004?x2?x??1,?2.

五、证明题

a2(a?1)2(a?2)2(a?3)2b21.证明:(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)2证明:

a2?a?1?2?a?2?2?a?3?2a22a?12a?32a?5a22a?122b2?b?1?2?b?2?2?b?3?2c4?c3b22b?12b?32b?5c4?c3b22b?122 c2?c?1?2?c?2?2?c?3?2c3?c2cc22c?12c?32c?5c2?c13?c2c22c?122d2?d?1?2?d?2?2?d?3?2d22d?12d?32d?5d22d?122推论40

a11a12?a1n2.设D????a,求证:D?D1?D2???Dn,其中Dkn?1,1an?1,2?an?1,n11?1将D中第k列元素换成x1,x2,,xn?1,1后所得的新行列式。

证明:将D增加一行和一列得到下列n?1阶行列式,此行列式显然为0。

1111a11a12a1nx1

an?1,1an?1,2an?1,nxn?11111n?1将此行列式按第一行展开,得???1?k?1A1k?0, k?1显然Akk?1???1?n?Dk,??1?k?1A1k???1?n?1Dk?k?1,2,,n?,

?k?1,2,,n?为

??1??n?1??1A1,?n?1??D,故D??Dk。

k?1n3.设a1,a2,?,an是数域P中互不相同的数,b1,b2,?,bn是数域P中任一组给定的数,用克拉默法则证明:有唯一的数域P上的多项式

f?x??c0?c1x?c2x2???cn?1xn?1 使f?ai??bi ?i?1,2,?,n?。

证明:由f?ai??bi得

?c0?c1a1?c2a12???cn?1a1n?1?b1?2n?1?c0?c1a2?c2a2???cn?1a2?b2 ?

.............................................??c?ca?ca2???can?1?b2nn?1nn?01n这是一个关于c0,c1,?,cn?1的线性方程组,且它的系数行列式为一个范德蒙行列式.由已知该行列式不为0,故线性方程组只有唯一解,即所求多项式是唯一的.