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)的(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})且(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}'})为((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|
]