復(fù)旦大學(xué)謝啟鴻高等代數(shù)每周一題[2021A04]參考解答

2021-10-27 18:12 作者:CharlesMa0606 0人讀過 | 我要投稿

本文是本人給出的2021年復(fù)旦大學(xué)謝啟鴻高等代數(shù)的每周一題[問題2021A04]的解答

題目來自于復(fù)旦大學(xué)謝啟鴻教授在他的博客提供的每周一題練習(xí)

（鏈接：https://www.cnblogs.com/torsor/p/15329047.html）

本文僅供學(xué)習(xí)交流，如有錯(cuò)誤懇請(qǐng)指正！

[問題2021A04]求下列行列式的值：

$%5Cleft%7CA%5Cright%7C%3D%5Cleft%7C%5Cbegin%7Bmatrix%7Da_1%5E2%2Bn-1%26a_1%2Ba_2%26a_1%2Ba_3%26%5Ccdots%26a_1%2Ba_n%5C%5Ca_1%2Ba_2%26a_2%5E2%2B1%261%26%5Ccdots%261%5C%5Ca_1%2Ba_3%261%26a_3%5E2%2B1%26%5Ccdots%261%5C%5C%5Cvdots%26%5Cvdots%26%5Cvdots%26%5C%20%26%5Cvdots%5C%5Ca_1%2Ba_n%261%261%26%5Ccdots%26a_n%5E2%2B1%5C%5C%5Cend%7Bmatrix%7D%5Cright%7C$

解（解法一，拆分、降階公式，計(jì)算量較大）

1°當(dāng) $a_i%5Cneq0%2C%5Cforall2%5Cle%20i%5Cle%20n$ 時(shí)我們對(duì) $%7CA%7C$ 的第一列進(jìn)行拆分，有：

$%5Cleft%7CA%5Cright%7C%3D%5Cleft%7C%5Cbegin%7Bmatrix%7Da_1%5E2%26a_1%2Ba_2%26a_1%2Ba_3%26%5Ccdots%26a_1%2Ba_n%5C%5Ca_1%2Ba_2%26a_2%5E2%2B1%261%26%5Ccdots%261%5C%5Ca_1%2Ba_3%261%26a_3%5E2%2B1%26%5Ccdots%261%5C%5C%5Cvdots%26%5Cvdots%26%5Cvdots%26%5C%20%26%5Cvdots%5C%5Ca_1%2Ba_n%261%261%26%5Ccdots%26a_n%5E2%2B1%5C%5C%5Cend%7Bmatrix%7D%5Cright%7C%2B%5Cleft%7C%5Cbegin%7Bmatrix%7Dn-1%26a_1%2Ba_2%26a_1%2Ba_3%26%5Ccdots%26a_1%2Ba_n%5C%5C0%26a_2%5E2%2B1%261%26%5Ccdots%261%5C%5C0%261%26a_3%5E2%2B1%26%5Ccdots%261%5C%5C%5Cvdots%26%5Cvdots%26%5Cvdots%26%5C%20%26%5Cvdots%5C%5C0%261%261%26%5Ccdots%26a_n%5E2%2B1%5C%5C%5Cend%7Bmatrix%7D%5Cright%7C.$

先計(jì)算前一項(xiàng)，將其記作 $%7CB%7C$ ，則：

$%5Cleft%7CB%5Cright%7C%3D%5Cleft%7C%5Cleft(%5Cbegin%7Bmatrix%7Da_1%5E2%26a_1%2Ba_2%26a_1%2Ba_3%26%5Ccdots%26a_1%2Ba_n%5C%5Ca_1%2Ba_2%26a_2%5E2%260%26%5Ccdots%260%5C%5Ca_1%2Ba_3%260%26a_3%5E2%26%5Ccdots%260%5C%5C%5Cvdots%26%5Cvdots%26%5Cvdots%26%5C%20%26%5Cvdots%5C%5Ca_1%2Ba_n%260%260%26%5Ccdots%26a_n%5E2%5C%5C%5Cend%7Bmatrix%7D%5Cright)-%5Cleft(%5Cbegin%7Bmatrix%7D0%5C%5C1%5C%5C1%5C%5C%5Cvdots%5C%5C1%5C%5C%5Cend%7Bmatrix%7D%5Cright)%5Cleft(-I_1%5Cright)%5Cleft(%5Cbegin%7Bmatrix%7D0%261%261%26%5Ccdots%261%5C%5C%5Cend%7Bmatrix%7D%5Cright)%5Cright%7C$

$%3D%5Cfrac%7B1%7D%7B%5Cleft%7C-I_1%5Cright%7C%7D%5Cleft%7C%5Cbegin%7Bmatrix%7Da_1%5E2%26a_1%2Ba_2%26a_1%2Ba_3%26%5Ccdots%26a_1%2Ba_n%5C%5Ca_1%2Ba_2%26a_2%5E2%260%26%5Ccdots%260%5C%5Ca_1%2Ba_3%260%26a_3%5E2%26%5Ccdots%260%5C%5C%5Cvdots%26%5Cvdots%26%5Cvdots%26%5C%20%26%5Cvdots%5C%5Ca_1%2Ba_n%260%260%26%5Ccdots%26a_n%5E2%5C%5C%5Cend%7Bmatrix%7D%5Cright%7C%5Cleft%7C-I_1-%5Cleft(%5Cbegin%7Bmatrix%7D0%261%261%26%5Ccdots%261%5C%5C%5Cend%7Bmatrix%7D%5Cright)%5Cleft(%5Cast%5Cright)%5E%7B-1%7D%5Cleft(%5Cbegin%7Bmatrix%7D0%5C%5C1%5C%5C1%5C%5C%5Cvdots%5C%5C1%5C%5C%5Cend%7Bmatrix%7D%5Cright)%5Cright%7C$

而利用遞推法可以求出

$%5Cleft%7C%5Cbegin%7Bmatrix%7Da_1%5E2%26a_1%2Ba_2%26a_1%2Ba_3%26%5Ccdots%26a_1%2Ba_n%5C%5Ca_1%2Ba_2%26a_2%5E2%260%26%5Ccdots%260%5C%5Ca_1%2Ba_3%260%26a_3%5E2%26%5Ccdots%260%5C%5C%5Cvdots%26%5Cvdots%26%5Cvdots%26%5C%20%26%5Cvdots%5C%5Ca_1%2Ba_n%260%260%26%5Ccdots%26a_n%5E2%5C%5C%5Cend%7Bmatrix%7D%5Cright%7C%3Da_1%5E2%5Ccdots%20a_n%5E2-%5Csum_%7Bi%3D2%7D%5E%7Bn%7D%7Ba_2%5E2%5Ccdots%5Cwidehat%7Ba_i%5E2%7D%5Cleft(a_1%2Ba_i%5Cright)%5E2%5Ccdots%20a_n%5E2%7D.$

注意到只需要求出逆陣的后 $n-1$ 行、后 $n-1$ 列元素的和即可,從而

$%5Cleft%7CB%5Cright%7C%3D%5Cleft(a_1%5E2-%5Csum_%7Bi%3D2%7D%5E%7Bn%7D%5Cfrac%7B%5Cleft(a_1%2Ba_i%5Cright)%5E2%7D%7Ba_i%5E2%7D%5Cright)%5Cleft(1%2B%5Csum_%7Bi%3D2%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Ba_i%5E2%7D%2B%5Cfrac%7B%5Cleft(%5Csum_%7Bi%3D2%7D%5E%7Bn%7D%5Cfrac%7Ba_1%2Ba_i%7D%7Ba_i%5E2%7D%5Cright)%5E2%7D%7Ba_1%5E2-%5Csum_%7Bi%3D2%7D%5E%7Bn%7D%5Cfrac%7B%5Cleft(a_1%2Ba_i%5Cright)%5E2%7D%7Ba_i%5E2%7D%7D%5Cright)%5Cprod_%7Bi%3D2%7D%5E%7Bn%7Da_i%5E2$

$%3D%5Cleft%5B%5Cleft(a_1%5E2-%5Csum_%7Bi%3D2%7D%5E%7Bn%7D%5Cfrac%7B%5Cleft(a_1%2Ba_i%5Cright)%5E2%7D%7Ba_i%5E2%7D%5Cright)%5Cleft(1%2B%5Csum_%7Bi%3D2%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Ba_i%5E2%7D%5Cright)%2B%5Cleft(%5Csum_%7Bi%3D2%7D%5E%7Bn%7D%5Cfrac%7Ba_1%2Ba_i%7D%7Ba_i%5E2%7D%5Cright)%5E2%5Cright%5D%5Cprod_%7Bi%3D2%7D%5E%7Bn%7Da_i%5E2$

$%3D%5Cleft%5Ba_1%5E2-2a_1%5Csum_%7Bi%3D2%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Ba_i%7D-%5Cleft(n-1%5Cright)-%5Cleft(%5Cleft(n-1%5Cright)%5Csum_%7Bi%3D2%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Ba_i%5E2%7D%5Cright)%2B%5Cleft(%5Csum_%7Bi%3D2%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Ba_i%7D%5Cright)%5E2%5Cright%5D%5Cprod_%7Bi%3D2%7D%5E%7Bn%7Da_i%5E2$

$%3D%5Cleft%5B%5Cleft(a_1-%5Csum_%7Bi%3D2%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Ba_i%7D%5Cright)%5E2-%5Cleft(n-1%5Cright)%5Cleft(1%2B%5Csum_%7Bi%3D2%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Ba_i%5E2%7D%5Cright)%5Cright%5D%5Cprod_%7Bi%3D2%7D%5E%7Bn%7Da_i%5E2$

再計(jì)算后一項(xiàng)，將其記作 $%7CC%7C$ ，有：

$%5Cleft%7CC%5Cright%7C%3D%5Cleft%7C%5Cbegin%7Bmatrix%7Dn-1%26a_1%2Ba_2%26a_1%2Ba_3%26%5Ccdots%26a_1%2Ba_n%5C%5C0%26a_2%5E2%2B1%261%26%5Ccdots%261%5C%5C0%261%26a_3%5E2%2B1%26%5Ccdots%261%5C%5C%5Cvdots%26%5Cvdots%26%5Cvdots%26%5C%20%26%5Cvdots%5C%5C0%261%261%26%5Ccdots%26a_n%5E2%2B1%5C%5C%5Cend%7Bmatrix%7D%5Cright%7C%3D%5Cleft(n-1%5Cright)%5Cleft%7C%5Cbegin%7Bmatrix%7Da_2%5E2%2B1%261%261%26%5Ccdots%261%5C%5C1%26a_3%5E2%2B1%261%26%5Ccdots%261%5C%5C1%261%26a_3%5E2%2B1%26%5Ccdots%261%5C%5C%5Cvdots%26%5Cvdots%26%5Cvdots%26%5C%20%26%5Cvdots%5C%5C1%261%261%26%5Ccdots%26a_n%5E2%2B1%5C%5C%5Cend%7Bmatrix%7D%5Cright%7C$

$%3D%5Cleft(n-1%5Cright)%5Cleft(%5Cleft%7C%5Cbegin%7Bmatrix%7Da_2%5E2%260%260%26%5Ccdots%260%5C%5C0%26a_3%5E2%260%26%5Ccdots%260%5C%5C0%260%26a_4%5E2%26%5Ccdots%260%5C%5C%5Cvdots%26%5Cvdots%26%5Cvdots%26%5C%20%26%5Cvdots%5C%5C0%260%260%26%5Ccdots%26a_n%5E2%5C%5C%5Cend%7Bmatrix%7D%5Cright%7C%2B%5Csum_%7Bi%2Cj%3D1%7D%5E%7Bn%7DC_%7Bij%7D%5Cright)%3D%5Cleft(n-1%5Cright)%5Cleft(1%2B%5Csum_%7Bi%3D2%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Ba_i%5E2%7D%5Cright)%5Cprod_%7Bi%3D2%7D%5E%7Bn%7Da_i%5E2$

從而

$%5Cleft%7CA%5Cright%7C%3D%5Cleft(a_1-%5Csum_%7Bi%3D2%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Ba_i%7D%5Cright)%5E2%5Cprod_%7Bi%3D2%7D%5E%7Bn%7Da_i%5E2.%5Cleft(a_i%5Cneq0%5Cright)$

2°當(dāng)有且只有一個(gè) $a_i%3D0%2C2%5Cle%20i%5Cle%20n$ 時(shí)我們可以做行對(duì)換和列對(duì)換把它換到右下角，從而不妨只研究 $a_n%3D0$ 的情況，其余情況同理，我們有：

$%5Cleft%7CA%5Cright%7C%3D%5Cleft%7C%5Cbegin%7Bmatrix%7Da_1%5E2%2Bn-1%26a_1%2Ba_2%26a_1%2Ba_3%26%5Ccdots%26a_1%5C%5Ca_1%2Ba_2%26a_2%5E2%2B1%261%26%5Ccdots%261%5C%5Ca_1%2Ba_3%261%26a_3%5E2%2B1%26%5Ccdots%261%5C%5C%5Cvdots%26%5Cvdots%26%5Cvdots%26%5C%20%26%5Cvdots%5C%5Ca_1%261%261%26%5Ccdots%261%5C%5C%5Cend%7Bmatrix%7D%5Cright%7C%3D%5Cleft%7C%5Cbegin%7Bmatrix%7Dn-1%26a_2%26a_3%26%5Ccdots%26a_%7Bn-1%7D%5C%5Ca_2%26a_2%5E2%260%26%5Ccdots%260%5C%5Ca_3%260%26a_3%5E2%26%5Ccdots%260%5C%5C%5Cvdots%26%5Cvdots%26%5Cvdots%26%5C%20%26%5Cvdots%5C%5Ca_%7Bn-1%7D%260%260%26%5Ccdots%26a_%7Bn-1%7D%5E2%5C%5C%5Cend%7Bmatrix%7D%5Cright%7C$

$%3D%5Cprod_%7Bi%3D2%7D%5E%7Bn-1%7Da_i%5E2%5Cleft%7C%5Cbegin%7Bmatrix%7Dn-1%261%261%26%5Ccdots%261%5C%5C1%261%260%26%5Ccdots%260%5C%5C1%260%261%26%5Ccdots%260%5C%5C%5Cvdots%26%5Cvdots%26%5Cvdots%26%5C%20%26%5Cvdots%5C%5C1%260%260%26%5Ccdots%261%5C%5C%5Cend%7Bmatrix%7D%5Cright%7C%3D%5Cprod_%7Bi%3D2%7D%5E%7Bn-1%7Da_i%5E2%5Cleft%7C%5Cbegin%7Bmatrix%7D1%261%261%26%5Ccdots%261%5C%5C0%261%260%26%5Ccdots%260%5C%5C0%260%261%26%5Ccdots%260%5C%5C%5Cvdots%26%5Cvdots%26%5Cvdots%26%5C%20%26%5Cvdots%5C%5C0%260%260%26%5Ccdots%261%5C%5C%5Cend%7Bmatrix%7D%5Cright%7C%3D%5Cprod_%7Bi%3D2%7D%5E%7Bn-1%7Da_i%5E2$

3°當(dāng)有不少于兩個(gè) $a_i%3D0%2C2%5Cle%20i%5Cle%20n$ 時(shí)，有兩行相同，從而 $%7CA%7C%3D0$ .

$%5BQ.E.D%5D$

（解法二，矩陣乘法，計(jì)算量非常小，但較難想到）

注意到

$A%3D%5Cleft(%5Cbegin%7Bmatrix%7Da_1%5E2%2Bn-1%26a_1%2Ba_2%26a_1%2Ba_3%26%5Ccdots%26a_1%2Ba_n%5C%5Ca_1%2Ba_2%26a_2%5E2%2B1%261%26%5Ccdots%261%5C%5Ca_1%2Ba_3%261%26a_3%5E2%2B1%26%5Ccdots%261%5C%5C%5Cvdots%26%5Cvdots%26%5Cvdots%26%5C%20%26%5Cvdots%5C%5Ca_1%2Ba_n%261%261%26%5Ccdots%26a_n%5E2%2B1%5C%5C%5Cend%7Bmatrix%7D%5Cright)%3D%5Cleft(%5Cbegin%7Bmatrix%7Da_1%261%261%26%5Ccdots%261%5C%5C1%26a_2%260%26%5Ccdots%260%5C%5C1%260%26a_3%26%5Ccdots%260%5C%5C%5Cvdots%26%5Cvdots%26%5Cvdots%26%5C%20%26%5Cvdots%5C%5C1%260%260%26%5Ccdots%26a_n%5C%5C%5Cend%7Bmatrix%7D%5Cright)%5E2$

從而由爪形行列式的相關(guān)結(jié)論，立即得到

$%5Cleft%7CA%5Cright%7C%3D%5Cleft(%5Cprod_%7Bi%3D1%7D%5E%7Bn%7Da_i-%5Csum_%7Bk%3D2%7D%5E%7Bn%7D%7Ba_2%5Ccdots%5Cwidehat%7Ba_k%7D%5Ccdots%20a_n%7D%5Cright)%5E2%3D%5Cleft(a_1-%5Csum_%7Bi%3D2%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Ba_i%7D%5Cright)%5E2%5Cprod_%7Bi%3D2%7D%5E%7Bn%7Da_i%5E2.$