关于高等代数的计算题II(行列式)

$\bf计算：$$\bf(02浙大二)$设${S_n} = {x_1}^k + {x_2}^k +  \cdots  + {x_n}^k\left( {k = 0,1,2, \cdots } \right),{a_{ij}} = {S_{i + j - 2}}\left( {i,j = 1,2, \cdots ,n} \right)$，计算\[D = \left| {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{12}}}& \cdots &{{a_{1n}}} \\ 
{{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2n}}} \\ 
\cdots & \cdots & \cdots & \cdots \\ 
{{a_{n1}}}&{{a_{n2}}}& \cdots &{{a_{nn}}} 
\end{array}} \right|\]

1

$\bf计算：$$\bf(11华师一)$

\[{D_n} = \left| {\begin{array}{*{20}{c}}
{2 + {x_1}}&{2 + x_1^2}& \cdots &{2 + x_1^n} \\
{2 + {x_2}}&{2 + x_2^2}& \cdots &{2 + x_2^n} \\
\vdots & \vdots & \vdots & \vdots \\
{2 + {x_n}}&{2 + x_n^2}& \cdots &{2 + x_n^n}
\end{array}} \right|\]

1

$\bf计算：$$\bf(07武大二)$

\[{D_n} = \left| {\begin{array}{*{20}{c}}
{{a_1} + b}&{{a_2}}& \cdots &{{a_n}} \\
{{a_1}}&{{a_2} + b}& \cdots &{{a_n}} \\
\vdots & \vdots & \ddots & \vdots \\
{{a_1}}&{{a_2}}& \cdots &{{a_n} + b}
\end{array}} \right|\]

1

$\bf计算：$$\bf(11华科一)$

\[{D_n} = \left| {\begin{array}{*{20}{c}}
1&{{a_1}}& \cdots &{{a_{n - 1}}} \\
{{b_1}}&{{x_1}}& \cdots &0 \\
\cdots & \cdots & \cdots & \cdots \\
{{b_{n - 1}}}&0& \cdots &{{x_{n - 1}}}
\end{array}} \right|\]

1

$\bf计算：$$\bf(09浙大二)$

\[{D_n} = \left| {\begin{array}{*{20}{c}}
{2{a_1}{b_1}}&{{a_1}{b_2} + {a_2}{b_1}}& \cdots &{{a_1}{b_n} + {a_n}{b_1}} \\
{{a_2}{b_1} + {a_1}{b_2}}&{2{a_2}{b_2}}& \cdots &{{a_2}{b_n} + {a_n}{b_2}} \\
\vdots & \vdots & \ddots & \vdots \\
{{a_n}{b_1} + {a_1}{b_n}}&{{a_n}{b_2} + {a_2}{b_n}}& \cdots &{2{a_n}{b_n}}
\end{array}} \right|\]

1

$\bf计算：$$\bf(12中科院三)$

\[{D_n} = \left| {\begin{array}{*{20}{c}}
{{a_1}^2}&{{a_1}{a_2} + 1}& \cdots &{{a_1}{a_n} + 1} \\
{{a_2}{a_1} + 1}&{{a_2}^2}& \cdots &{{a_2}{a_n} + 1} \\
\cdots & \cdots & \ddots & \cdots \\
{{a_n}{a_1} + 1}&{{a_n}{a_2} + 1}& \cdots &{{a_n}^2}
\end{array}} \right|\]

1

$\bf计算：$$\bf(10中科院一)$

\[{D_n} = \left| {\begin{array}{*{20}{c}}
{1 + {a_1} + {x_1}}&{{a_1} + {x_2}}& \cdots &{{a_1} + {x_n}} \\
{{a_2} + {x_1}}&{1 + {a_2} + {x_2}}&{}&{{a_2} + {x_n}} \\
\vdots & \vdots & \ddots & \vdots \\
{{a_n} + {x_1}}&{{a_n} + {x_2}}&{}&{1 + {a_n} + {x_n}}
\end{array}} \right|\]

1

$\bf计算：$$\bf(04南开一)$设$n$阶反对称阵$A=(a_{ij})$的行列式为1，对任意的$x$，计算$B=(a_{ij}+x)$的行列式

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$\bf计算：$