许以超书上一行列式求解 - Eufisky - The lost book
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2011年南开大学高等代数试题

许以超书上一行列式求解

Eufisky posted @ 2015年9月26日 16:17 in 高等代数 with tags 许以超 , 1126 阅读

许以超第二版书上P73页留了个行列式求解的思考题,还是比较棘手的,下面给出自己的解答.


\[{\Delta _n} = \det \left( {\begin{array}{*{20}{c}}{1 + {x_1}{y_1}}&{1 + {x_1}{y_2}}& \cdots &{1 + {x_1}{y_n}}\\{1 + {x_1}y_1^2}&{1 + {x_1}y_2^2}& \cdots &{1 + {x_1}y_n^2}\\\vdots & \vdots &{}& \vdots \\{1 + {x_1}y_1^n}&{1 + {x_1}y_2^n}& \cdots &{1 + {x_1}y_n^n}\end{array}} \right).\]

解:先来个引理(许以超自己给出的).


事实上,我们有

\begin{align*}&\det A + x\sum\limits_{j,k = 1}^n {{A_{jk}}} = \det \left( {\begin{array}{*{20}{c}}{{a_{11}} + x}& \cdots &{{a_{1n}} + x}\\\vdots &{}& \vdots \\{{a_{n1}} + x}& \cdots &{{a_{nn}} + x}\end{array}} \right)\\=& \det A + x\det \left( {\begin{array}{*{20}{c}}1&1& \cdots &1&1\\{{a_{21}} - {a_{11}}}&{{a_{22}} - {a_{12}}}& \cdots &{{a_{2,n - 1}} - {a_{1,n - 1}}}&{{a_{2n}} - {a_{1n}}}\\{{a_{31}} - {a_{21}}}&{{a_{32}} - {a_{22}}}& \cdots &{{a_{3,n - 1}} - {a_{2,n - 1}}}&{{a_{3n}} - {a_{2n}}}\\\vdots & \vdots &{}& \vdots & \vdots \\{{a_{n1}} - {a_{n - 1,1}}}&{{a_{n2}} - {a_{n - 1,2}}}& \cdots &{{a_{n,n - 1}} - {a_{n - 1,n - 1}}}&{{a_{nn}} - {a_{n - 1,n}}}\end{array}} \right).\end{align*}

其中$A_{kj}$是方阵$A=(a_{jk})$的第$k$行,第$j$列位置的元素的代数余子式.


因此

\begin{align*}&{\Delta _n} = \det \left( {\begin{array}{*{20}{c}}{1 + {x_1}{y_1}}&{1 + {x_1}{y_2}}& \cdots &{1 + {x_1}{y_n}}\\{1 + {x_1}y_1^2}&{1 + {x_1}y_2^2}& \cdots &{1 + {x_1}y_n^2}\\\vdots & \vdots &{}& \vdots \\{1 + {x_1}y_1^n}&{1 + {x_1}y_2^n}& \cdots &{1 + {x_1}y_n^n}\end{array}} \right)\\= &\det A + \det \left( {\begin{array}{*{20}{c}}1&1& \cdots &1\\{{x_1}\left( {y_1^2 - {y_1}} \right)}&{{x_1}\left( {y_2^2 - {y_2}} \right)}& \cdots &{x_1\left( {y_n^2 - {y_n}} \right)}\\\vdots & \vdots &{}& \vdots \\{{x_1}\left( {y_1^n - y_1^{n - 1}} \right)}&{{x_1}\left( {y_2^n - y_2^{n - 1}} \right)}& \cdots &{{x_1}\left( {y_n^n - y_n^{n - 1}} \right)}\end{array}} \right)\\=& \det A + x_1^{n - 1}\det \left( {\begin{array}{*{20}{c}}1&1& \cdots &1\\{y_1^2 - {y_1}}&{y_2^2 - {y_2}}& \cdots &{y_n^2 - {y_n}}\\\vdots & \vdots &{}& \vdots \\{y_1^n - y_1^{n - 1}}&{y_2^n - y_2^{n - 1}}& \cdots &{y_n^n - y_n^{n - 1}}\end{array}} \right).\end{align*}

 

对上面的行列式进行升阶:

\[\det \left( {\begin{array}{*{20}{c}}1&1& \cdots &1\\{y_1^2 - {y_1}}&{y_2^2 - {y_2}}& \cdots &{y_n^2 - {y_n}}\\\vdots & \vdots &{}& \vdots \\{y_1^n - y_1^{n - 1}}&{y_2^n - y_2^{n - 1}}& \cdots &{y_n^n - y_n^{n - 1}}\end{array}} \right) = \det \left( {\begin{array}{*{20}{c}}1&{{y_1}}&{{y_2}}& \cdots &{{y_n}}\\0&1&1& \cdots &1\\0&{y_1^2 - {y_1}}&{y_2^2 - {y_2}}& \cdots &{y_n^2 - {y_n}}\\\vdots & \vdots & \vdots &{}& \vdots \\0&{y_1^n - y_1^{n - 1}}&{y_2^n - y_2^{n - 1}}& \cdots &{y_n^n - y_n^{n - 1}}\end{array}} \right).\]

将第一行加到第三行,第三行加到第四行,$\cdots$,最后将第$n$行加到第$n+1$行:

\begin{align*}\det \left( {\begin{array}{*{20}{c}}1&{{y_1}}&{{y_2}}& \cdots &{{y_n}}\\0&1&1& \cdots &1\\1&{y_1^2}&{y_2^2}& \cdots &{y_n^2}\\\vdots & \vdots & \vdots &{}& \vdots \\1&{y_1^n}&{y_2^n}& \cdots &{y_n^n}\end{array}} \right) &= \det \left( {\begin{array}{*{20}{c}}1&{{y_1}}&{{y_2}}& \cdots &{{y_n}}\\1&1&1& \cdots &1\\1&{y_1^2}&{y_2^2}& \cdots &{y_n^2}\\\vdots & \vdots & \vdots &{}& \vdots \\1&{y_1^n}&{y_2^n}& \cdots &{y_n^n}\end{array}} \right) + \det \left( {\begin{array}{*{20}{c}}1&{{y_1}}&{{y_2}}& \cdots &{{y_n}}\\{ - 1}&0&0& \cdots &0\\1&{y_1^2}&{y_2^2}& \cdots &{y_n^2}\\\vdots & \vdots & \vdots &{}& \vdots \\1&{y_1^n}&{y_2^n}& \cdots &{y_n^n}\end{array}} \right)\\& = \left( { - 1} \right) \cdot \prod\limits_{k = 1}^n {\left( {{y_k} - 1} \right)} \cdot \prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} + \prod\limits_{k = 1}^n {{y_k}} \cdot \prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} \\& = \prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} \cdot \left[ { - \prod\limits_{k = 1}^n {\left( {{y_k} - 1} \right)} + \prod\limits_{k = 1}^n {{y_k}} } \right].\end{align*}

因此

\begin{align*}{\Delta _n} &= \det A + x_1^{n - 1}\prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} \cdot \left[ { - \prod\limits_{k = 1}^n {\left( {{y_k} - 1} \right)} + \prod\limits_{k = 1}^n {{y_k}} } \right]\\& = x_1^n \cdot \prod\limits_{k = 1}^n {{y_k}} \cdot \prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} + x_1^{n - 1}\prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} \cdot \left[ { - \prod\limits_{k = 1}^n {\left( {{y_k} - 1} \right)} + \prod\limits_{k = 1}^n {{y_k}} } \right]\\& = x_1^{n - 1}\prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} \cdot \left[ { - \prod\limits_{k = 1}^n {\left( {{y_k} - 1} \right)} + \left( {{x_1} + 1} \right)\prod\limits_{k = 1}^n {{y_k}} } \right].\end{align*}


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