文章目录
- @[toc]
- 第一章:行列式
- 第一节|方程组与行列式
- 二元线性方程组和二阶行列式
- 第二节| n n n阶行列式
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- 第三节|行列式的性质和计算
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文章目录
- @[toc]
- 第一章:行列式
- 第一节|方程组与行列式
- 二元线性方程组和二阶行列式
- 第二节| n n n阶行列式
- 第三节|行列式的性质和计算
第一章:行列式
第一节|方程组与行列式
二元线性方程组和二阶行列式
- 二元线性方程组
{ a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 22 x 2 = b 2 (1) \begin{cases} a_{11} x_{1} + a_{12} x_{2} = b_{1} \\ a_{21} x_{1} + a_{22} x_{2} = b_{2} \end{cases} \tag{1} {a11x1+a12x2=b1a21x1+a22x2=b2(1)
- 二阶行列式
D = ∣ a 11 a 12 a 21 a 22 ∣ = a 11 a 22 − a 12 a 21 D = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11} a_{22} - a_{12} a_{21} D= a11a21a12a22 =a11a22−a12a21
- 二阶行列式 D D D称为二元线性方程组( 1 1 1)的系数行列式
第二节| n n n阶行列式
排列
- 由 1 1 1, 2 2 2, ⋯ \cdots ⋯, n n n组成的一个有序数组称为一个 n n n级排列
逆序数
- 在一个排列中,如果一对数的前后位置与大小顺序相反,则称它们为一个逆序
- 一个排列中所有逆序的总数称为该排列的逆序数
- 逆序数为偶数的排列称为偶排列,逆序数为奇数的排列称为奇排列
- n n n级排列 i 1 i 2 ⋯ i n i_{1} i_{2} \cdots i_{n} i1i2⋯in的逆序数记为 τ ( i 1 i 2 ⋯ i n ) \tau(i_{1} i_{2} \cdots i_{n}) τ(i1i2⋯in)
对换
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将一个排列中某两个数的位置互换,而其余的数不动,就得到另一个排列,这样一个变换称为一个对换
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对换改变排列的奇偶性
- 证明
- 先考虑相邻两个数的对换,设排列 ⋯ j k ⋯ \cdots jk \cdots ⋯jk⋯经 j j j, k k k对换变成排列 ⋯ k j ⋯ \cdots kj \cdots ⋯kj⋯,排列中除 j j j, k k k两个数本身顺序改变外,其他数的顺序并没有改变,而 j j j, k k k之间,若 j < k j < k j<k,则经过对换后的排列的逆序数比原排列的逆序数增加 1 1 1,若 j > k j > k j>k,则经对换后的排列的逆序数比原排列的逆序数减少 1 1 1,因此,对换 j j j, k k k的位置改变排列的奇偶性
- 再看一般情况,设排列 ⋯ j i 1 i 2 ⋯ i m k ⋯ \cdots j i_{1} i_{2} \cdots i_{m} k \cdots ⋯ji1i2⋯imk⋯,经 j j j, k k k对换变成排列 ⋯ k i 1 i 2 ⋯ i m j ⋯ \cdots k i_{1} i_{2} \cdots i_{m} j \cdots ⋯ki1i2⋯imj⋯,先对原排列施行 m m m次相邻两个数的对换,原排列变为 ⋯ j k i 1 i 2 ⋯ i m ⋯ \cdots jk i_{1} i_{2} \cdots i_{m} \cdots ⋯jki1i2⋯im⋯,再经过 m + 1 m + 1 m+1次相邻两个数的对换,原排列变为 ⋯ k i 1 i 2 ⋯ i m j ⋯ \cdots k i_{1} i_{2} \cdots i_{m} j \cdots ⋯ki1i2⋯imj⋯,因为相邻两个数的对换改变排列的奇偶性,共施行了 2 m + 1 2m + 1 2m+1次相邻两个数的对换,于是改变了排列的奇偶性
- 证明
-
在 n ( ≥ 2 ) n (\geq 2) n(≥2)级排列中,奇偶排列各占一半,即各有 n ! 2 \cfrac{n!}{2} 2n!个
例题 1 1 1
- 问题:求排列 n ( n − 1 ) ⋯ 3 2 1 n \ (n - 1) \cdots 3 \ 2 \ 1 n (n−1)⋯3 2 1的逆序数,并讨论其奇偶性
- 解答
- τ ( n ( n − 1 ) ⋯ 3 2 1 ) = ( n − 1 ) + ( n − 2 ) + ⋯ + 3 + 2 + 1 = n ( n − 1 ) 2 \tau(n \ (n - 1) \cdots 3 \ 2 \ 1) = (n - 1) + (n - 2) + \cdots + 3 + 2 + 1 = \cfrac{n (n - 1)}{2} τ(n (n−1)⋯3 2 1)=(n−1)+(n−2)+⋯+3+2+1=2n(n−1),当 n = 4 k n = 4k n=4k, 4 k + 1 4k + 1 4k+1时,该排列是偶排列,当 n = 4 k + 2 n = 4k + 2 n=4k+2, 4 k + 3 4k + 3 4k+3时,该排列是奇排列
n n n阶行列式
D = ∑ ( j 1 j 2 ⋯ j n ) ( − 1 ) τ ( j 1 j 2 ⋯ j n ) a 1 j 1 a 2 j 2 ⋯ a n j n D = \sum\limits_{(j_{1} j_{2} \cdots j_{n})}{(-1)^{\tau(j_{1} j_{2} \cdots j_{n})} a_{1 j_{1}} a_{2 j_{2}} \cdots a_{n j_{n}}} D=(j1j2⋯jn)∑(−1)τ(j1j2⋯jn)a1j1a2j2⋯anjn
- 也可写成
D = ∑ ( j 1 j 2 ⋯ j n ) ( − 1 ) τ ( i 1 i 2 ⋯ i n ) + τ ( j 1 j 2 ⋯ j n ) a i 1 j 1 a i 2 j 2 ⋯ a i n j n D = \sum\limits_{(j_{1} j_{2} \cdots j_{n})}{(-1)^{\tau(i_{1} i_{2} \cdots i_{n}) + \tau(j_{1} j_{2} \cdots j_{n})} a_{i_{1} j_{1}} a_{i_{2} j_{2}} \cdots a_{i_{n} j_{n}}} D=(j1j2⋯jn)∑(−1)τ(i1i2⋯in)+τ(j1j2⋯jn)ai1j1ai2j2⋯ainjn
-
- 证明
- 将项 a i 1 j 1 a i 2 j 2 ⋯ a i n j n a_{i_{1} j_{1}} a_{i_{2} j_{2}} \cdots a_{i_{n} j_{n}} ai1j1ai2j2⋯ainjn经一系列元素对换排成 a 1 j 1 ′ a 2 j 2 ′ ⋯ a n j n ′ a_{1 j_{1}^{'}} a_{2 j_{2}^{'}} \cdots a_{n j_{n}^{'}} a1j1′a2j2′⋯anjn′,每作一次元素对换,相应的行下标和列下标所成排列也作了一次对换,因此逆序数之和奇偶性不变,于是
- 证明
∑ ( j 1 j 2 ⋯ j n ) ( − 1 ) τ ( i 1 i 2 ⋯ i n ) + τ ( j 1 j 2 ⋯ j n ) a i 1 j 1 a i 2 j 2 ⋯ a i n j n = ∑ ( j 1 ′ j 2 ′ ⋯ j n ′ ) ( − 1 ) τ ( j 1 ′ j 2 ′ ⋯ j n ′ ) a 1 j 1 ′ a 2 j 2 ′ ⋯ a n j n ′ \sum\limits_{(j_{1} j_{2} \cdots j_{n})}{(-1)^{\tau(i_{1} i_{2} \cdots i_{n}) + \tau(j_{1} j_{2} \cdots j_{n})} a_{i_{1} j_{1}} a_{i_{2} j_{2}} \cdots a_{i_{n} j_{n}}} = \sum\limits_{(j_{1}^{'} j_{2}^{'} \cdots j_{n}^{'})}{(-1)^{\tau(j_{1}^{'} j_{2}^{'} \cdots j_{n}^{'})} a_{1 j_{1}^{'}} a_{2 j_{2}^{'}} \cdots a_{n j_{n}^{'}}} (j1j2⋯jn)∑(−1)τ(i1i2⋯in)+τ(j1j2⋯jn)ai1j1ai2j2⋯ainjn=(j1′j2′⋯jn′)∑(−1)τ(j1′j2′⋯jn′)a1j1′a2j2′⋯anjn′
第三节|行列式的性质和计算
性质 1 1 1
D = ∣ a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ = D T = ∣ a 11 a 21 ⋯ a n 1 a 12 a 22 ⋯ a n 2 ⋮ ⋮ ⋮ a 1 n a 2 n ⋯ a n n ∣ D= \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} = D^{T} = \begin{vmatrix} a_{11} & a_{21} & \cdots & a_{n1} \\ a_{12} & a_{22} & \cdots & a_{n2} \\ \vdots & \vdots & & \vdots \\ a_{1n} & a_{2n} & \cdots & a_{nn} \\ \end{vmatrix} D= a11a21⋮an1a12a22⋮an2⋯⋯⋯a1na2n⋮ann =DT= a11a12⋮a1na21a22⋮a2n⋯⋯⋯an1an2⋮ann
- 证明
- 在 D D D中位于第 i i i行,第 j j j列的元素 a i j a_{ij} aij在 D T D^{T} DT中位于第 j j j行,第 i i i列
D T = ∣ b 11 b 12 ⋯ b 1 n b 21 b 22 ⋯ b 2 n ⋮ ⋮ ⋮ b n 1 b n 2 ⋯ b n n ∣ D^{T} = \begin{vmatrix} b_{11} & b_{12} & \cdots & b_{1n} \\ b_{21} & b_{22} & \cdots & b_{2n} \\ \vdots & \vdots & & \vdots \\ b_{n1} & b_{n2} & \cdots & b_{nn} \\ \end{vmatrix} DT= b11b21⋮bn1b12b22⋮bn2⋯⋯⋯b1nb2n⋮bnn
-
- 则有 b i j = a j i b_{ij} = a_{ji} bij=aji( i i i, j = 1 j = 1 j=1, 2 2 2, ⋯ \cdots ⋯, n n n)
D T = ∑ ( j 1 j 2 ⋯ j n ) ( − 1 ) τ ( j 1 j 2 ⋯ j n ) b 1 j 1 b 2 j 2 ⋯ b n j n = ∑ ( j 1 j 2 ⋯ j n ) ( − 1 ) τ ( j 1 j 2 ⋯ j n ) a j 1 1 a j 2 2 ⋯ a j n n = D \begin{aligned} D^{T} &= \sum_{(j_{1}j_{2} \cdots j_{n})}(-1)^{\tau(j_{1}j_{2} \cdots j_{n})}b_{1j_{1}}b_{2j_{2}} \cdots b_{nj_{n}} \\ &= \sum_{(j_{1}j_{2} \cdots j_{n})}(-1)^{\tau(j_{1}j_{2} \cdots j_{n})}a_{j_{1}1}a_{j_{2}2} \cdots a_{j_{n}n} \\ &= D \end{aligned} DT=(j1j2⋯jn)∑(−1)τ(j1j2⋯jn)b1j1b2j2⋯bnjn=(j1j2⋯jn)∑(−1)τ(j1j2⋯jn)aj11aj22⋯ajnn=D
性质 2 2 2:互换 n n n阶行列式的任意两行(列),行列式仅改变符号
- 证明
- 设行列式
D = ∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ a i 1 a i 2 ⋯ a i n ⋮ ⋮ ⋮ a k 1 a k 2 ⋯ a k n ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ D = \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & & \vdots \\ a_{i1} & a_{i2} & \cdots & a_{in} \\ \vdots & \vdots & & \vdots \\ a_{k1} & a_{k2} & \cdots & a_{kn} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} D= a11⋮ai1⋮ak1⋮an1a12⋮ai2⋮ak2⋮an2⋯⋯⋯⋯a1n⋮ain⋮akn⋮ann
-
- 交换 D D D的第 i i i行和第 k k k行,得行列式
D 1 = ∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ a k 1 a k 2 ⋯ a k n ⋮ ⋮ ⋮ a i 1 a i 2 ⋯ a i n ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ D_{1} = \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & & \vdots \\ a_{k1} & a_{k2} & \cdots & a_{kn} \\ \vdots & \vdots & & \vdots \\ a_{i1} & a_{i2} & \cdots & a_{in} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} D1= a11⋮ak1⋮ai1⋮an1a12⋮ak2⋮ai2⋮an2⋯⋯⋯⋯a1n⋮akn⋮ain⋮ann
-
- D D D中任意一项 a 1 j 1 ⋯ a i j i ⋯ a k j k ⋯ a n j n a_{1j_{1}} \cdots a_{ij_{i}} \cdots a_{kj_{k}} \cdots a_{nj_{n}} a1j1⋯aiji⋯akjk⋯anjn也是 D 1 D_{1} D1中的一项 a 1 j 1 ⋯ a k j k ⋯ a i j i ⋯ a n j n a_{1j_{1}} \cdots a_{kj_{k}} \cdots a_{ij_{i}} \cdots a_{nj_{n}} a1j1⋯akjk⋯aiji⋯anjn,其中 a k j k a_{kj_{k}} akjk是 D 1 D_{1} D1中第 i i i行第 j k j_{k} jk列元素, a i j i a_{ij_{i}} aiji是 D 1 D_{1} D1中第 k k k行第 j i j_{i} ji列元素
- 项 a 1 j 1 ⋯ a i j i ⋯ a k j k ⋯ a n j n a_{1j_{1}} \cdots a_{ij_{i}} \cdots a_{kj_{k}} \cdots a_{nj_{n}} a1j1⋯aiji⋯akjk⋯anjn在 D D D中的符号为 ( − 1 ) τ ( j 1 ⋯ j i ⋯ j k ⋯ j n ) (-1)^{\tau(j_{1} \cdots j_{i} \cdots j_{k} \cdots j_{n})} (−1)τ(j1⋯ji⋯jk⋯jn),项 a 1 j 1 ⋯ a k j k ⋯ a i j i ⋯ a n j n a_{1j_{1}} \cdots a_{kj_{k}} \cdots a_{ij_{i}} \cdots a_{nj_{n}} a1j1⋯akjk⋯aiji⋯anjn在 D 1 D_{1} D1中的符号为 ( − 1 ) τ ( j 1 ⋯ j k ⋯ j i ⋯ j n ) (-1)^{\tau(j_{1} \cdots j_{k} \cdots j_{i} \cdots j_{n})} (−1)τ(j1⋯jk⋯ji⋯jn)
( − 1 ) τ ( j 1 ⋯ j i ⋯ j k ⋯ j n ) = − ( − 1 ) τ ( j 1 ⋯ j k ⋯ j i ⋯ j n ) (-1)^{\tau(j_{1} \cdots j_{i} \cdots j_{k} \cdots j_{n})} = -(-1)^{\tau(j_{1} \cdots j_{k} \cdots j_{i} \cdots j_{n})} (−1)τ(j1⋯ji⋯jk⋯jn)=−(−1)τ(j1⋯jk⋯ji⋯jn)
-
- 因此 D = − D 1 D = -D_{1} D=−D1
推论 1 1 1:若行列式中某两行(列)的元素对应相等,则行列式为零
性质 3 3 3
∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ k a i 1 k a i 2 ⋯ k a i n ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ = k ∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ a i 1 a i 2 ⋯ a i n ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & & \vdots \\ ka_{i1} & ka_{i2} & \cdots & ka_{in} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} = k \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & & \vdots \\ a_{i1} & a_{i2} & \cdots & a_{in} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} a11⋮kai1⋮an1a12⋮kai2⋮an2⋯⋯⋯a1n⋮kain⋮ann =k a11⋮ai1⋮an1a12⋮ai2⋮an2⋯⋯⋯a1n⋮ain⋮ann
- 证明
D = ∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ a i 1 a i 2 ⋯ a i n ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ , D 1 = ∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ k a i 1 k a i 2 ⋯ k a i n ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ D = \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & & \vdots \\ a_{i1} & a_{i2} & \cdots & a_{in} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} , D_{1} = \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & & \vdots \\ ka_{i1} & ka_{i2} & \cdots & ka_{in} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} D= a11⋮ai1⋮an1a12⋮ai2⋮an2⋯⋯⋯a1n⋮ain⋮ann ,D1= a11⋮kai1⋮an1a12⋮kai2⋮an2⋯⋯⋯a1n⋮kain⋮ann
D 1 = ∑ ( j 1 ⋯ j i ⋯ j n ) ( − 1 ) τ ( j 1 ⋯ j i ⋯ j n ) a 1 j 1 ⋯ k a i j i ⋯ a n j n = k ∑ ( j 1 ⋯ j i ⋯ j n ) ( − 1 ) τ ( j 1 ⋯ j i ⋯ j n ) a 1 j 1 ⋯ a i j i ⋯ a n j n = k D \begin{aligned} D_{1} &= \sum_{(j_{1} \cdots j_{i} \cdots j_{n})}(-1)^{\tau(j_{1} \cdots j_{i} \cdots j_{n})}a_{1j_{1}} \cdots ka_{ij_{i}} \cdots a_{nj_{n}} \\ &= k\sum_{(j_{1} \cdots j_{i} \cdots j_{n})}(-1)^{\tau(j_{1} \cdots j_{i} \cdots j_{n})}a_{1j_{1}} \cdots a_{ij_{i}} \cdots a_{nj_{n}} \\ &= kD \end{aligned} D1=(j1⋯ji⋯jn)∑(−1)τ(j1⋯ji⋯jn)a1j1⋯kaiji⋯anjn=k(j1⋯ji⋯jn)∑(−1)τ(j1⋯ji⋯jn)a1j1⋯aiji⋯anjn=kD
推论 2 2 2:若行列式的两行(列)的元素对应成比例,则该行列式为 0 0 0
性质 4 4 4
∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ a i 1 + a i 1 ′ a i 2 + a i 2 ′ ⋯ a i n + a i n ′ ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ = ∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ a i 1 a i 2 ⋯ a i n ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ + ∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ a i 1 ′ a i 2 ′ ⋯ a i n ′ ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & & \vdots \\ a_{i1} + a_{i1}^{'} & a_{i2} + a_{i2}^{'} & \cdots & a_{in} + a_{in}^{'} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} = \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & & \vdots \\ a_{i1} & a_{i2} & \cdots & a_{in} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} + \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & & \vdots \\ a_{i1}^{'} & a_{i2}^{'} & \cdots & a_{in}^{'} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} a11⋮ai1+ai1′⋮an1a12⋮ai2+ai2′⋮an2⋯⋯⋯a1n⋮ain+ain′⋮ann = a11⋮ai1⋮an1a12⋮ai2⋮an2⋯⋯⋯a1n⋮ain⋮ann + a11⋮ai1′⋮an1a12⋮ai2′⋮an2⋯⋯⋯a1n⋮ain′⋮ann
- 证明
D = ∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ a i 1 + a i 1 ′ a i 2 + a i 2 ′ ⋯ a i n + a i n ′ ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ D = \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & & \vdots \\ a_{i1} + a_{i1}^{'} & a_{i2} + a_{i2}^{'} & \cdots & a_{in} + a_{in}^{'} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} D= a11⋮ai1+ai1′⋮an1a12⋮ai2+ai2′⋮an2⋯⋯⋯a1n⋮ain+ain′⋮ann
D 1 = ∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ a i 1 a i 2 ⋯ a i n ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ , D 2 = ∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ a i 1 ′ a i 2 ′ ⋯ a i n ′ ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ D_{1} = \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & & \vdots \\ a_{i1} & a_{i2} & \cdots & a_{in} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} , D_{2} = \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & & \vdots \\ a_{i1}^{'} & a_{i2}^{'} & \cdots & a_{in}^{'} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} D1= a11⋮ai1⋮an1a12⋮ai2⋮an2⋯⋯⋯a1n⋮ain⋮ann ,D2= a11⋮ai1′⋮an1a12⋮ai2′⋮an2⋯⋯⋯a1n⋮ain′⋮ann
D = ∑ ( j 1 ⋯ j i ⋯ j n ) ( − 1 ) τ ( j 1 ⋯ j i ⋯ j n ) a 1 j 1 ⋯ ( a i j i + a i j i ′ ) ⋯ a n j n = ∑ ( j 1 ⋯ j i ⋯ j n ) ( − 1 ) τ ( j 1 ⋯ j i ⋯ j n ) a 1 j 1 ⋯ a i j i ⋯ a n j n + ∑ ( j 1 ⋯ j i ⋯ j n ) ( − 1 ) τ ( j 1 ⋯ j i ⋯ j n ) a 1 j 1 ⋯ a i j i ′ ⋯ a n j n = D 1 + D 2 \begin{aligned} D &= \sum_{(j_{1} \cdots j_{i} \cdots j_{n})}(-1)^{\tau(j_{1} \cdots j_{i} \cdots j_{n})}a_{1j_{1}} \cdots (a_{ij_{i}} + a_{ij_{i}}^{'}) \cdots a_{nj_{n}} \\ &= \sum_{(j_{1} \cdots j_{i} \cdots j_{n})}(-1)^{\tau(j_{1} \cdots j_{i} \cdots j_{n})}a_{1j_{1}} \cdots a_{ij_{i}} \cdots a_{nj_{n}} + \sum_{(j_{1} \cdots j_{i} \cdots j_{n})}(-1)^{\tau(j_{1} \cdots j_{i} \cdots j_{n})}a_{1j_{1}} \cdots a_{ij_{i}}^{'} \cdots a_{nj_{n}} \\ &= D_{1} + D_{2} \end{aligned} D=(j1⋯ji⋯jn)∑(−1)τ(j1⋯ji⋯jn)a1j1⋯(aiji+aiji′)⋯anjn=(j1⋯ji⋯jn)∑(−1)τ(j1⋯ji⋯jn)a1j1⋯aiji⋯anjn+(j1⋯ji⋯jn)∑(−1)τ(j1⋯ji⋯jn)a1j1⋯aiji′⋯anjn=D1+D2
性质 5 5 5:把行列式的某行(列)的各元素乘 k k k后加到另一行(列)的对应元素上,行列式的值不变
例题 1 1 1
- 问题:计算 n n n阶行列式
∣ a b b ⋯ b b a b ⋯ b b b a ⋯ b ⋮ ⋮ ⋮ ⋮ b b b ⋯ a ∣ \begin{vmatrix} a & b & b & \cdots & b \\ b & a & b & \cdots & b \\ b & b & a & \cdots & b \\ \vdots & \vdots & \vdots & & \vdots \\ b & b & b & \cdots & a \end{vmatrix} abb⋮bbab⋮bbba⋮b⋯⋯⋯⋯bbb⋮a
- 解法一:
∣ a b b ⋯ b b a b ⋯ b b b a ⋯ b ⋮ ⋮ ⋮ ⋮ b b b ⋯ a ∣ = ∣ a + ( n − 1 ) b b b ⋯ b a + ( n − 1 ) b a b ⋯ b a + ( n − 1 ) b b a ⋯ b ⋮ ⋮ ⋮ ⋮ a + ( n − 1 ) b b b ⋯ a ∣ = [ a + ( n − 1 ) b ] ∣ 1 b b ⋯ b 0 a − b 0 ⋯ 0 0 0 a − b ⋯ 0 ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ a − b ∣ = [ a + ( n − 1 ) b ] ( a − b ) n − 1 \begin{vmatrix} a & b & b & \cdots & b \\ b & a & b & \cdots & b \\ b & b & a & \cdots & b \\ \vdots & \vdots & \vdots & & \vdots \\ b & b & b & \cdots & a \end{vmatrix} = \begin{vmatrix} a + (n - 1)b & b & b & \cdots & b \\ a + (n - 1)b & a & b & \cdots & b \\ a + (n - 1)b & b & a & \cdots & b \\ \vdots & \vdots & \vdots & & \vdots \\ a + (n - 1)b & b & b & \cdots & a \end{vmatrix} = [a + (n - 1)b] \begin{vmatrix} 1 & b & b & \cdots & b \\ 0 & a - b & 0 & \cdots & 0 \\ 0 & 0 & a - b & \cdots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \cdots & a - b \end{vmatrix} = [a + (n - 1)b](a - b)^{n - 1} abb⋮bbab⋮bbba⋮b⋯⋯⋯⋯bbb⋮a = a+(n−1)ba+(n−1)ba+(n−1)b⋮a+(n−1)bbab⋮bbba⋮b⋯⋯⋯⋯bbb⋮a =[a+(n−1)b] 100⋮0ba−b0⋮0b0a−b⋮0⋯⋯⋯⋯b00⋮a−b =[a+(n−1)b](a−b)n−1
- 解法二:
∣ a b b ⋯ b b a b ⋯ b b b a ⋯ b ⋮ ⋮ ⋮ ⋮ b b b ⋯ a ∣ = ∣ a b b ⋯ b b − a a − b 0 ⋯ 0 b − a 0 a − b ⋯ 0 ⋮ ⋮ ⋮ ⋮ b − a 0 0 ⋯ a − b ∣ = ∣ a + ( n − 1 ) b b b ⋯ b 0 a − b 0 ⋯ 0 0 0 a − b ⋯ 0 ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ a − b ∣ = [ a + ( n − 1 ) b ] ( a − b ) n − 1 \begin{vmatrix} a & b & b & \cdots & b \\ b & a & b & \cdots & b \\ b & b & a & \cdots & b \\ \vdots & \vdots & \vdots & & \vdots \\ b & b & b & \cdots & a \end{vmatrix} = \begin{vmatrix} a & b & b & \cdots & b \\ b - a & a - b & 0 & \cdots & 0 \\ b - a & 0 & a - b & \cdots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ b - a & 0 & 0 & \cdots & a - b \end{vmatrix} = \begin{vmatrix} a + (n - 1)b & b & b & \cdots & b \\ 0 & a - b & 0 & \cdots & 0 \\ 0 & 0 & a - b & \cdots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \cdots & a - b \end{vmatrix} = [a + (n - 1)b](a - b)^{n - 1} abb⋮bbab⋮bbba⋮b⋯⋯⋯⋯bbb⋮a = ab−ab−a⋮b−aba−b0⋮0b0a−b⋮0⋯⋯⋯⋯b00⋮a−b = a+(n−1)b00⋮0ba−b0⋮0b0a−b⋮0⋯⋯⋯⋯b00⋮a−b =[a+(n−1)b](a−b)n−1
对称行列式和反对称行列式
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在行列式中,若 a i j = a j i a_{ij} = a_{ji} aij=aji,则称 D D D为对称行列式
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在行列式中,若 a i j = − a j i a_{ij} = -a_{ji} aij=−aji,则称 D D D为反对称行列式
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反对称行列式中主对角线上的元素等于 0 0 0
例题 2 2 2
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问题:证明奇数阶反对称行列式的值等于 0 0 0
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解答
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- 设 D D D是反对称行列式, D D D可以写成
D = ∣ 0 a 12 a 13 ⋯ a 1 n − a 12 0 a 23 ⋯ a 2 n − a 13 − a 23 0 ⋯ a 3 n ⋮ ⋮ ⋮ ⋮ − a 1 n − a 2 n − a 3 n ⋯ 0 ∣ D = \begin{vmatrix} 0 & a_{12} & a_{13} & \cdots & a_{1n} \\ -a_{12} & 0 & a_{23} & \cdots & a_{2n} \\ -a_{13} & -a_{23} & 0 & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & & \vdots \\ -a_{1n} & -a_{2n} & -a_{3n} & \cdots & 0 \end{vmatrix} D= 0−a12−a13⋮−a1na120−a23⋮−a2na13a230⋮−a3n⋯⋯⋯⋯a1na2na3n⋮0
D = D T = ∣ 0 − a 12 − a 13 ⋯ − a 1 n a 12 0 − a 23 ⋯ − a 2 n a 13 a 23 0 ⋯ − a 3 n ⋮ ⋮ ⋮ ⋮ a 1 n a 2 n a 3 n ⋯ 0 ∣ = ( − 1 ) n ∣ 0 a 12 a 13 ⋯ a 1 n − a 12 0 a 23 ⋯ a 2 n − a 13 − a 23 0 ⋯ a 3 n ⋮ ⋮ ⋮ ⋮ − a 1 n − a 2 n − a 3 n ⋯ 0 ∣ = ( − 1 ) n D D = D^{T} = \begin{vmatrix} 0 & -a_{12} & -a_{13} & \cdots & -a_{1n} \\ a_{12} & 0 & -a_{23} & \cdots & -a_{2n} \\ a_{13} & a_{23} & 0 & \cdots & -a_{3n} \\ \vdots & \vdots & \vdots & & \vdots \\ a_{1n} & a_{2n} & a_{3n} & \cdots & 0 \end{vmatrix} = (-1)^{n} \begin{vmatrix} 0 & a_{12} & a_{13} & \cdots & a_{1n} \\ -a_{12} & 0 & a_{23} & \cdots & a_{2n} \\ -a_{13} & -a_{23} & 0 & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & & \vdots \\ -a_{1n} & -a_{2n} & -a_{3n} & \cdots & 0 \end{vmatrix} = (-1)^{n}D D=DT= 0a12a13⋮a1n−a120a23⋮a2n−a13−a230⋮a3n⋯⋯⋯⋯−a1n−a2n−a3n⋮0 =(−1)n 0−a12−a13⋮−a1na120−a23⋮−a2na13a230⋮−a3n⋯⋯⋯⋯a1na2na3n⋮0 =(−1)nD
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- 因 n n n为奇数,故 D = 0 D = 0 D=0