性质
性质一
∣ A O O B ∣ = ∣ A ∗ O B ∣ = ∣ A O ∗ B ∣ = ∣ A ∣ ⋅ ∣ B ∣ \left|\begin{array}{ll} \boldsymbol{A} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{B} \end{array}\right|=\left|\begin{array}{ll} \boldsymbol{A} & * \\ \boldsymbol{O} & \boldsymbol{B} \end{array}\right|=\left|\begin{array}{cc} \boldsymbol{A} & \boldsymbol{O} \\ *& \boldsymbol{B} \end{array}\right|=|\boldsymbol{A}| \cdot|\boldsymbol{B}| ∣∣∣∣AOOB∣∣∣∣=∣∣∣∣AO∗B∣∣∣∣=∣∣∣∣A∗OB∣∣∣∣=∣A∣⋅∣B∣
性质二
设 A , B \boldsymbol{A}, \boldsymbol{B} A,B分别是m阶和n阶矩阵,则
∣ O A B O ∣ = ( − 1 ) m n ∣ A ∣ ⋅ ∣ B ∣ \left|\begin{array}{ll} \boldsymbol{O} & \boldsymbol{A} \\ \boldsymbol{B} & \boldsymbol{O} \end{array}\right|=(-1)^{m n}|\boldsymbol{A}| \cdot|\boldsymbol{B}| ∣∣∣∣OBAO∣∣∣∣=(−1)mn∣A∣⋅∣B∣
证明
证明一
对于性质一，我们只证
∣ A O ∗ B ∣ = ∣ A ∣ ⋅ ∣ B ∣ \left|\begin{array}{cc} \boldsymbol{A} & \boldsymbol{O} \\ *& \boldsymbol{B} \end{array}\right|=|\boldsymbol{A}| \cdot|\boldsymbol{B}| ∣∣∣∣A∗OB∣∣∣∣=∣A∣⋅∣B∣
其他同理可得
设
D = ∣ a 11 ⋯ a 1 k ⋮ ⋮ 0 a k 1 ⋯ a k k c 11 ⋯ c 1 k b 11 ⋯ b 1 n ⋮ ⋮ ⋮ ⋮ c n 1 ⋯ c n k b n 1 ⋯ b n n ∣ D=\left|\begin{array}{cccccc} a_{11} & \cdots & a_{1 k} & & & \\ \vdots & & \vdots & & 0 & \\ a_{k 1} & \cdots & a_{k k} & & & \\ c_{11} & \cdots & c_{1 k} & b_{11} & \cdots & b_{1 n} \\ \vdots & & \vdots & \vdots & & \vdots \\ c_{n 1} & \cdots & c_{n k} & b_{n 1} & \cdots & b_{n n} \end{array}\right| D=∣∣∣∣∣∣∣∣∣∣∣∣∣∣a11⋮ak1c11⋮cn1⋯⋯⋯⋯a1k⋮akkc1k⋮cnkb11⋮bn10⋯⋯b1n⋮bnn∣∣∣∣∣∣∣∣∣∣∣∣∣∣
D 1 = ∣ a 11 ⋯ a 1 k ⋮ ⋮ a k 1 ⋯ a k k ∣ D_{1}=\left|\begin{array}{ccc} a_{11} & \cdots & a_{1 k} \\ \vdots & & \vdots \\ a_{k 1} & \cdots & a_{k k} \end{array}\right| D1=∣∣∣∣∣∣∣a11⋮ak1⋯⋯a1k⋮akk∣∣∣∣∣∣∣
D 2 = ∣ b 11 ⋯ b 1 n ⋮ ⋮ b n 1 ⋯ b n n ∣ D_{2}=\left|\begin{array}{ccc} b_{11} & \cdots & b_{1 n} \\ \vdots & & \vdots \\ b_{n 1} & \cdots & b_{n n} \end{array}\right| D2=∣∣∣∣∣∣∣b11⋮bn1⋯⋯b1n⋮bnn∣∣∣∣∣∣∣
通过行变换把 D 1 D_{1} D1化成下三角行列式
D 1 = ∣ p 11 0 ⋮ ⋱ p k 1 ⋯ p k k ∣ = p 11 ⋯ p k k D_{1}=\left|\begin{array}{ccc} p_{11} & & 0 \\ \vdots & \ddots & \\ p_{k 1} & \cdots & p_{k k} \end{array}\right|=p_{11} \cdots p_{k k} D1=∣∣∣∣∣∣∣p11⋮pk1⋱⋯0pkk∣∣∣∣∣∣∣=p11⋯pkk
同理，通过列变换把 D 2 D_{2} D2也化成下三角行列式
D 2 = ∣ q 11 0 ⋮ ⋱ q n 1 ⋯ q n n ∣ = q 11 ⋯ q n n D_{2}=\left|\begin{array}{ccc} q_{11} & & 0 \\ \vdots & \ddots & \\ q_{n 1} & \cdots & q_{n n} \end{array}\right|=q_{11} \cdots q_{n n} D2=∣∣∣∣∣∣∣q11⋮qn1⋱⋯0qnn∣∣∣∣∣∣∣=q11⋯qnn
对行列式 D D D的前 k k k行进行行变换，后n列进行列变换成
D = ∣ p 11 ⋮ ⋱ 0 p k 1 ⋯ p k k c 11 ⋯ c 1 k q 11 ⋮ ⋮ ⋮ ⋱ c n 1 ⋯ c n k q n 1 ⋯ q u n ∣ D=\left|\begin{array}{cccccc} p_{11} & & & & & \\ \vdots & \ddots & & & 0 & \\ p_{k 1} & \cdots & p_{k k} & & & \\ c_{11} & \cdots & c_{1 k} & q_{11} & & \\ \vdots & & \vdots & \vdots & \ddots & \\ c_{n 1} & \cdots & c_{n k} & q_{n 1} & \cdots & q_{u n} \end{array}\right| D=∣∣∣∣∣∣∣∣∣∣∣∣∣∣p11⋮pk1c11⋮cn1⋱⋯⋯⋯pkkc1k⋮cnkq11⋮qn10⋱⋯qun∣∣∣∣∣∣∣∣∣∣∣∣∣∣
由下三角行列式的性质可得
D = D 1 D 2 D=D_{1} D_{2} D=D1D2
证明二
设想将 ∣ O A B O ∣ \left|\begin{array}{ll}\boldsymbol{O} & \boldsymbol{A} \\ \boldsymbol{B} & \boldsymbol{O}\end{array}\right| ∣∣∣∣OBAO∣∣∣∣变为 k ⋅ ∣ B O O A ∣ k \cdot \left|\begin{array}{ll}\boldsymbol{B} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{A}\end{array}\right| k⋅∣∣∣∣BOOA∣∣∣∣的形式，然后用性质一即可解决。
用冒泡排序的思想将 B \boldsymbol{B} B中的每一行依次与 A \boldsymbol{A} A中的每一行交换，即可将 B \boldsymbol{B} B冒上去，这样总共进行了 m n mn mn次交换，由行列式的性质可得，行列式中任意交换两行要变号，所以
∣ O A B O ∣ = ( − 1 ) m n ∣ B O O A ∣ = ( − 1 ) m n ∣ A ∣ ⋅ ∣ B ∣ \left|\begin{array}{ll} \boldsymbol{O} & \boldsymbol{A} \\ \boldsymbol{B} & \boldsymbol{O} \end{array}\right|=(-1)^{m n}\left|\begin{array}{ll}\boldsymbol{B} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{A}\end{array}\right|=(-1)^{m n}|\boldsymbol{A}| \cdot|\boldsymbol{B}| ∣∣∣∣OBAO∣∣∣∣=(−1)mn∣∣∣∣BOOA∣∣∣∣=(−1)mn∣A∣⋅∣B∣
补充：各种排序算法的动画演示
参考文献
[1]同济大学数学系. 工程数学线性代数[M]. 高等教育出版社，2018
[2]朱灵 毕道旺. 线性代数[M]. 北京邮电出版社，2019
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[4]https://blog.csdn.net/xiaoxiaojie12321/article/details/81380834