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  • 关于高等代数的计算题II(行列式)

    $f计算:$$f(02浙大二)$设${S_n} = {x_1}^k + {x_2}^k +  cdots  + {x_n}^kleft( {k = 0,1,2, cdots } ight),{a_{ij}} = {S_{i + j - 2}}left( {i,j = 1,2, cdots ,n} ight)$,计算[D = left| {egin{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}} ight|]

    1

    $f计算:$$f(11华师一)$

    [{D_n} = left| {egin{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}} ight|]

    1

    $f计算:$$f(07武大二)$

    [{D_n} = left| {egin{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}} ight|]

    1

    $f计算:$$f(11华科一)$

    [{D_n} = left| {egin{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}} ight|]

    1

    $f计算:$$f(09浙大二)$

    [{D_n} = left| {egin{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}} ight|]

    1

    $f计算:$$f(12中科院三)$

    [{D_n} = left| {egin{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}} ight|]

    1

    $f计算:$$f(10中科院一)$

    [{D_n} = left| {egin{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}} ight|]

    1

    $f计算:$$f(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

    1   2

    $f计算:$

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  • 原文地址:https://www.cnblogs.com/ly142857/p/3747076.html
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