一个行列式题目

计算行列式:
\begin{vmatrix}
1+x_1 & 1+x_1^2 & \cdots & 1+x_1^n \\
1+x_2 & 1+x_2^2 & \cdots & 1+x_2^n \\
\vdots & \vdots & & \vdots \\
1+x_n & 1+x_n^2 & \cdots & 1+x_n^n
\end{vmatrix}
已邀请:
我这里给出一个更复杂的,来源于许以超的书。由于你这个转置后就是我的特殊情形,你可以类似地写出解答!
-----------------------------------------------------------------------------------------------

\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).\end{align*}
-----------------------------------------------------------------------------------------------
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:
\begin{align*}\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).\end{align*}
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*}
$\begin{vmatrix}
1+x_1 & 1+x_1^2 & \cdots & 1+x_1^n\\
1+x_2 & 1+x_2^2 & \cdots & 1+x_2^n\\
\vdots & \vdots & &\vdots\\
1+x_n & 1+x_n^2 & \cdots & 1+x_n^n
\end{vmatrix}
=
\begin{vmatrix}
1 & 1 & 1 & \cdots & 1\\
0 & 1+x_1 & 1+x_1^2 & \cdots & 1+x_1^n\\
0 & 1+x_2 & 1+x_2^2 & \cdots & 1+x_2^n\\
\vdots & \vdots & \vdots & &\vdots\\
0 & 1+x_n & 1+x_n^2 & \cdots & 1+x_n^n
\end{vmatrix}
\\
=
\begin{vmatrix}
1 & 1 & 1 & \cdots & 1\\
-1 & x_1 & x_1^2 & \cdots & x_1^n\\
-1 & x_2 & x_2^2 & \cdots & x_2^n\\
\vdots & \vdots & \vdots & &\vdots\\
-1 & x_n & x_n^2 & \cdots & x_n^n
\end{vmatrix}
\\
=
\begin{vmatrix}
2 & 1 & 1 & \cdots & 1\\
0 & x_1 & x_1^2 & \cdots & x_1^n\\
0 & x_2 & x_2^2 & \cdots & x_2^n\\
\vdots & \vdots & \vdots & &\vdots\\
0 & x_n & x_n^2 & \cdots & x_n^n
\end{vmatrix}
-
\begin{vmatrix}
1 & 1 & 1 & \cdots & 1\\
1 & x_1 & x_1^2 & \cdots & x_1^n\\
1 & x_2 & x_2^2 & \cdots & x_2^n\\
\vdots & \vdots & \vdots & &\vdots\\
1 & x_n & x_n^2 & \cdots & x_n^n
\end{vmatrix}
\\
=
2\prod_\limits{i=1}^{n} x_i \prod_\limits{1\le i<j\le n} (x_j-x_i)-\prod_\limits{i=1}^n (x_i-1)\prod_\limits{1\le i<j\le n}(x_j-x_i)
\\
=\left(2\prod_\limits{i=1}^{n} x_i-\prod_\limits{i=1}^n (x_i-1)\right)\prod_\limits{1\le i<j\le n}(x_j-x_i)$

要回复问题请先登录注册