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  • 2.6 拉普拉斯定理

    定义 1:
    (n)级矩阵(A)中任意取定(k)行,(k)列((1 leq k < n)),位于这些行和列交叉处的(k^2)个元素按原来的排法组成的(k)级矩阵的行列式称为(A)的一个(k)阶子式。取定(A)(i_1,i_2,dots ,i_k)行((i_1<i_2<dots <i_k))及(j_1,j_2,dots ,j_k)列((j_1<j_2<dots <j_k)),所得的(k)阶子式记为(A binom{i_1i_2dots i_k}{j_1j_2dots j_k}) ((1))

    划去这个(k)阶子式,剩下的元素按原来的排法组成((n-k))级矩阵,称为((1))的余子式。也是(A)的一个((n-k))阶子式,令({i_1'i_2'dots i_{n-k}'} = {1,2,dots ,n}) ({i_1i_2dots i_k})({j_1'j_2'dots j_{n-k}'} = {1,2,dots ,n}) ({j_1j_2dots j_k})(i_1'<i_2'<dots <i_{n-k}')(j_1'<j_2'<dots <j_{n-k}'),记余子式为(A binom{i_1'i_2'dots i_{n-k}'}{j_1'j_2'dots j_{n-k}'}) ((2))
    ((-1)^{(i_1 + i_2 + dots + i_k) + (j_1 + j_2 + dots + j_k)}A binom{i_1'i_2'dots i_{n-k}'}{j_1'j_2'dots j_{n-k}'})((1))的代数余子式。

    定理 1:
    (Laplace定理)在(n)级矩阵(A = (a_{ij})),取定第(i_1,i_2,dots ,i_k)行((i_1<i_2<dots <i_k)),则这(k)阶行元素形成的所有(k)阶子式与其代数余子式之和等于(|A|),即

    [|A| = sum_{1leq j_1<j_2<dots <j_k leq n} A binom{i_1i_2dots i_k}{j_1j_2dots j_k}(-1)^{(i_1 + i_2 + dots + i_k) + (j_1 + j_2 + dots + j_k)}A binom{i_1'i_2'dots i_{n-k}'}{j_1'j_2'dots j_{n-k}'} ]

    证明:

    [egin{aligned} |A| &= sum_{u_1dots u_kv_1dots v_{n-k}}(-1)^{ au(i_1i_2dots i_ki_1'i_2'dots i_{n-k}')+ au(u_1dots u_kv_1dots v_{n-k})}a_{i_1u_1}a_{i_2u_2}dots a_{i_ku_k}a_{i_1'v_1}a_{i_2'v_2}dots a_{i_{n-k}'v_{n-k}} \ &= sum_{u_1dots u_kv_1dots v_{n-k}}(-1)^{(i_1-1)+(i_2-2)+dots +(i_k-k)+ au(u_1dots u_kv_1dots v_{n-k})}a_{i_1u_1}a_{i_2u_2}dots a_{i_ku_k}a_{i_1'v_1}a_{i_2'v_2}dots a_{i_{n-k}'v_{n-k}} \ &= sum_{1leq j_1<j_2<dots <j_k leq n}sum_{u_1dots u_k}sum_{v_1dots v_{n-k}}(-1)^{(i_1+i_2+dots +i_k)-frac{k(k+1)}{2}}(-1)^{ au(u_1dots u_kv_1dots v_{n-k})}a_{i_1u_1}a_{i_2u_2}dots a_{i_ku_k}a_{i_1'v_1}a_{i_2'v_2}dots a_{i_{n-k}'v_{n-k}} end{aligned} ]

    [这一步三个连加号分别指的是选定k列、对k元排列求和、对n-k元排列求和 \ 考察(-1)^{ au(u_1dots u_kv_1dots v_{n-k})},\ u_1dots u_kv_1dots v_{n-k}经过s次对换变成j_1j_2dots j_kv_1dots v_{n-k} \ 因此(-1)^{ au(u_1dots u_kv_1dots v_{n-k})} = (-1)^s(-1)^{ au(j_1j_2dots j_kv_1dots v_{n-k})} \ 其中j_1j_2dots j_k的逆序数为0,则s = au(u_1dots u_k),故 ]

    [egin{aligned} (-1)^{ au(u_1dots u_kv_1dots v_{n-k})}&=(-1)^{ au(u_1dots u_k)}(-1)^{ au(j_1j_2dots j_kv_1dots v_{n-k})} \ &=(-1)^{ au(u_1dots u_k)}(-1)^{(j_1-1)+(j_2-2)+dots +(j_k-k)}(-1)^{ au(v_1dots v_{n-k})} end{aligned} ]

    [egin{aligned} |A|&= sum_{1leq j_1<j_2<dots <j_k leq n}sum_{u_1dots u_k}sum_{v_1dots v_{n-k}}(-1)^{(i_1+i_2+dots +i_k)-frac{k(k+1)}{2}}(-1)^{ au(u_1dots u_k)}(-1)^{(j_1-1)+(j_2-2)+dots +(j_k-k)}(-1)^{ au(v_1dots v_{n-k})}a_{i_1u_1}a_{i_2u_2}dots a_{i_ku_k}a_{i_1'v_1}a_{i_2'v_2}dots a_{i_{n-k}'v_{n-k}} \ &= sum_{1leq j_1<j_2<dots <j_k leq n}(-1)^{(i_1+i_2+dots +i_k)+(j_1+j_2+dots +j_k)}sum_{u_1dots u_k}(-1)^{ au(u_1dots u_k)}a_{i_1u_1}a_{i_2u_2}dots a_{i_ku_k}(sum_{v_1dots v_{n-k}}(-1)^{ au(v_1dots v_{n-k})}a_{i_1'v_1}a_{i_2'v_2}dots a_{i_{n-k}'v_{n-k}}) \ &= sum_{1leq j_1<j_2<dots <j_k leq n} A binom{i_1i_2dots i_k}{j_1j_2dots j_k}(-1)^{(i_1 + i_2 + dots + i_k) + (j_1 + j_2 + dots + j_k)}A binom{i_1'i_2'dots i_{n-k}'}{j_1'j_2'dots j_{n-k}'} end{aligned} ]

    推论:

    [egin{vmatrix} a_{11} & cdots & a_{1k} & 0 & cdots & 0 \ vdots & & vdots & vdots & & vdots \ a_{k1} & cdots & a_{kk} & 0 & cdots & 0 \ c_{11} & cdots & c_{1k} & b_{11} & cdots & b_{1r} \ vdots & & vdots & vdots & & vdots \ c_{r1} & cdots & c_{rk} & b_{r1} & cdots & b_{rr} end{vmatrix} = egin{vmatrix} a_{11} & cdots & a_{1k} \ vdots & & vdots \ a_{k1} & cdots & a_{kk} end{vmatrix} cdot egin{vmatrix} b_{11} & cdots & b_{1r} \ vdots & & vdots \ b_{r1} & cdots & b_{rr} end{vmatrix} cdot (-1)^{frac{k(k+1)}{2}} cdot (-1)^{frac{k(k+1)}{2}} ]

    即:

    [egin{vmatrix} A & 0 \ C & B end{vmatrix} = |A| cdot |B| ]

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  • 原文地址:https://www.cnblogs.com/miraclepbc/p/14443187.html
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