行列式的性质
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D=\left| \begin{matrix} a_{11} \quad a_{12} \quad \cdots \quad a_{1n} \\ a_{21} \quad a_{22} \quad \cdots \quad a_{2n} \\ \vdots \quad \vdots \quad \cdots \quad \vdots \\ a_{n1} \quad a_{n2} \quad \cdots \quad a_{nn} \end{matrix} \right|
D=∣∣∣∣∣∣∣∣∣a11a12⋯a1na21a22⋯a2n⋮⋮⋯⋮an1an2⋯ann∣∣∣∣∣∣∣∣∣
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D^T=\left| \begin{matrix} a_{11} \quad a_{21} \quad \cdots \quad a_{n1} \\ a_{12} \quad a_{22} \quad \cdots \quad a_{n2} \\ \vdots \quad \vdots \quad \cdots \quad \vdots \\ a_{1n} \quad a_{2n} \quad \cdots \quad a_{nn} \end{matrix} \right|
DT=∣∣∣∣∣∣∣∣∣a11a21⋯an1a12a22⋯an2⋮⋮⋯⋮a1na2n⋯ann∣∣∣∣∣∣∣∣∣
性质1:
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在行列式中, 行和列的位置是对称的, 对行成立的, 对列也成立.因此下面只介绍关于行列式的行的性质 .
性质2: 互换两行, 行列式变号.即
$\left| \begin{matrix} a_{11} \quad a_{12} \quad \cdots \quad a_{1n} \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{i1} \quad a_{i2} \quad \cdots \quad a_{in} \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{j1} \quad a_{j2} \quad \cdots \quad a_{jn} \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{n1} \quad a_{n2} \quad \cdots \quad a_{nn} \end{matrix} \right|= -\left| \begin{matrix} a_{11} \quad a_{12} \quad \cdots \quad a_{1n} \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{j1} \quad a_{j2} \quad \cdots \quad a_{jn} \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{i1} \quad a_{i2} \quad \cdots \quad a_{in} \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{n1} \quad a_{n2} \quad \cdots \quad a_{nn} \end{matrix} \right| $
思考如何证明?
考虑三阶行列式交换2行的结果.
$\left| \begin{matrix} a_{11} \quad a_{12} \quad a_{13} \ a_{21} \quad a_{22} \quad a_{23} \ a_{31} \quad a_{32} \quad a_{33} \end{matrix} \right| \rightarrow \left| \begin{matrix} a_{21} \quad a_{22} \quad a_{23} \ a_{11} \quad a_{12} \quad a_{13} \ a_{31} \quad a_{32} \quad a_{33} \end{matrix} \right| $
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=a_{21}a_{12}a_{33} +a_{11}a_{32}a_{23} + a_{22}a_{13}a_{31} - a_{31}a_{12}a_{23} - a_{32}a_{13}a_{21} - a_{11}a_{22}a_{33}
=a21a12a33+a11a32a23+a22a13a31−a31a12a23−a32a13a21−a11a22a33
原始的行列式
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= a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{21}a_{32}a_{13} - a_{31}a_{22}a_{13} - a_{32}a_{23}a_{11} - a_{33}a_{21}a_{12}
=a11a22a33+a12a23a31+a21a32a13−a31a22a13−a32a23a11−a33a21a12
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对原式按照第一行展开a_{11}(-1)^(1 + 1)\left| \begin{matrix} a_{22} \quad a_{23} \\ a_{32} \quad a_{33}\end{matrix} \right| + a_{12}(-1)^(1 + 2)\left| \begin{matrix} a_{21} \quad a_{23} \\ a_{31} \quad a_{33}\end{matrix} \right| + a_{13}(-1)^(1 + 3)\left| \begin{matrix} a_{21} \quad a_{22} \\ a_{31} \quad a_{32}\end{matrix} \right|
对原式按照第一行展开a11(−1)(1+1)∣∣∣∣a22a23a32a33∣∣∣∣+a12(−1)(1+2)∣∣∣∣a21a23a31a33∣∣∣∣+a13(−1)(1+3)∣∣∣∣a21a22a31a32∣∣∣∣
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对交换之后的矩阵按照第二行展开 a_{11}(-1)^(2 + 1)\left| \begin{matrix} a_{22} \quad a_{23} \\ a_{32} \quad a_{33}\end{matrix} \right| + a_{12}(-1)^(2 + 2)\left| \begin{matrix} a_{21} \quad a_{23} \\ a_{31} \quad a_{33}\end{matrix} \right| + a_{13}(-1)^(2 + 3)\left| \begin{matrix} a_{21} \quad a_{22} \\ a_{31} \quad a_{32}\end{matrix} \right|
对交换之后的矩阵按照第二行展开a11(−1)(2+1)∣∣∣∣a22a23a32a33∣∣∣∣+a12(−1)(2+2)∣∣∣∣a21a23a31a33∣∣∣∣+a13(−1)(2+3)∣∣∣∣a21a22a31a32∣∣∣∣
推论: 若行列式中有两行元素完全相同, 则行列式为0.
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|D| = -|D| \Rightarrow 0
∣D∣=−∣D∣⇒0
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A_{ij}为元素a_{ij}的代数余子式, 则有
Aij为元素aij的代数余子式,则有
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a_{j1}A_{i1} +a_{j2}A_{i2}+ \cdots + a_{jn}A_{in} = 0 (i \neq j)
aj1Ai1+aj2Ai2+⋯+ajnAin=0(i=j)
怎么证明,考虑下面行列式
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\left| \begin{matrix} a_{11} \quad a_{12} \quad \cdots \quad a_{1n} \\ \vdots \quad \vdots \quad \cdots \quad \vdots \\ a_{i1} \quad a_{i2} \quad \cdots \quad a_{in} \\ \vdots \quad \vdots \quad \cdots \quad \vdots \\ a_{i1} \quad a_{i2} \quad \cdots \quad a_{in} \\ \vdots \quad \vdots \quad \cdots \quad \vdots \\ a_{n1} \quad a_{n2} \quad \cdots \quad a_{nn} \end{matrix} \right|
∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣a11a12⋯a1n⋮⋮⋯⋮ai1ai2⋯ain⋮⋮⋯⋮ai1ai2⋯ain⋮⋮⋯⋮an1an2⋯ann∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣
这个行列式第i行和第j行的元素相同, 根据第i行展开可得
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D=a_{i1}A_{i1} + a_{i2}A_{i2} + \cdots + a_{in}A_{in}
D=ai1Ai1+ai2Ai2+⋯+ainAin根据性质2的推论, 行列式中有两行元素完全相同,行列式的值为0, 所以
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D=0, 又因为第i行和第j行的元素相同, 所以
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a_{i1} = a_{j1}, a_{i2}=a{j2} \cdots a_{in}=a_{jn}
ai1=aj1,ai2=aj2⋯ain=ajn带入到上面的式子得到
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D=a_{j1}A_{i1} + a_{j2}A_{i2} + \cdots + a_{jn}A_{in}=0
D=aj1Ai1+aj2Ai2+⋯+ajnAin=0
即行列式中某一行(第j行)的元素与第i行元素的的代数余子式乘积之和是0, 同理列也一样.
可得
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a_{j1}A_{i1} +a_{j2}A_{i2}+ \cdots + a_{jn}A_{in} = 0 (i \neq j)
aj1Ai1+aj2Ai2+⋯+ajnAin=0(i=j)
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a_{j1}A_{i1} +a_{j2}A_{i2}+ \cdots + a_{jn}A_{in} = D (i = j)
aj1Ai1+aj2Ai2+⋯+ajnAin=D(i=j)
性质3: 用数k乘行列式中某一行的所有元素, 等于用k乘此行列式.即:
$\left| \begin{matrix} a_{11} \quad a_{12} \quad \cdots \quad a_{1n} \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{i1} \quad a_{i2} \quad \cdots \quad a_{in} \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{n1} \quad a_{n2} \quad \cdots \quad a_{nn} \end{matrix} \right|= k\left| \begin{matrix} a_{11} \quad a_{12} \quad \cdots \quad a_{1n} \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{i1} \quad a_{i2} \quad \cdots \quad a_{in} \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{n1} \quad a_{n2} \quad \cdots \quad a_{nn} \end{matrix} \right| $
推论: 某一行的所有元素的公因子可以提到行列式符号的外面.
比如
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\left| \begin{matrix} -8 \quad 4 \quad -6 \\ 2 \quad 1 \quad -1 \\ 16 \quad -2 \quad 7 \end{matrix}\right| = 2 \left| \begin{matrix} -4 \quad 2 \quad -3 \\ 2 \quad 1 \quad -1 \\ 16 \quad -2 \quad 7 \end{matrix}\right|= 4 \left| \begin{matrix} -2 \quad 2 \quad -3 \\ 1 \quad 1 \quad -1 \\ 8 \quad -2 \quad 7 \end{matrix}\right|
∣∣∣∣∣∣−84−621−116−27∣∣∣∣∣∣=2∣∣∣∣∣∣−42−321−116−27∣∣∣∣∣∣=4∣∣∣∣∣∣−22−311−18−27∣∣∣∣∣∣
性质4: 行列式某一行元素加上另一行对应元素的k倍, 行列式的值不变.即:
$\left| \begin{matrix} a_{11} \quad a_{12} \quad \cdots \quad a_{1n} \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{i1} + ka_{j1} \quad a_{i2} \quad \cdots \quad a_{in} + ka_{jn} \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{j1} \quad a_{j2} \quad \cdots \quad a_{jn} \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{n1} \quad a_{n2} \quad \cdots \quad a_{nn} \end{matrix} \right|= \left| \begin{matrix} a_{11} \quad a_{12} \quad \cdots \quad a_{1n} \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{i1} \quad a_{i2} \quad \cdots \quad a_{in} \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{j1} \quad a_{j2} \quad \cdots \quad a_{jn} \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{n1} \quad a_{n2} \quad \cdots \quad a_{nn} \end{matrix} \right| $
根据第i行展开
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(a_{i1} + ka_{j1})A_{i1} + (a_{i2} + ka_{j2})A_{i2} + \cdots + (a_{in} + ka_{jn})A_{in}
(ai1+kaj1)Ai1+(ai2+kaj2)Ai2+⋯+(ain+kajn)Ain
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展开:a_{i1}A_{i1} + a_{i2}A_{i2} + \cdots + a_{in}A_{in} +ka_{j1}A_{i1} + \cdots + ka_{jn}A_{in}
展开:ai1Ai1+ai2Ai2+⋯+ainAin+kaj1Ai1+⋯+kajnAin
根
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a_{j1}A_{i1} +a_{j2}A_{i2}+ \cdots + a_{jn}A_{in} = 0 (i \neq j)得到结论
aj1Ai1+aj2Ai2+⋯+ajnAin=0(i=j)得到结论
性质5: 若行列式某一行的元素是两数之和,则行列式可拆成两个行列式的和.即:
$\left| \begin{matrix} a_{11} \quad a_{12} \quad \cdots \quad a_{1n} \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{i1} + b_1 \quad a_{i2}+ b_2 \quad \cdots \quad a_{in}+b_n \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{n1} \quad a_{n2} \quad \cdots \quad a_{nn} \end{matrix} \right|= \left| \begin{matrix} a_{11} \quad a_{12} \quad \cdots \quad a_{1n} \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{i1} \quad a_{i2} \quad \cdots \quad a_{in} \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{n1} \quad a_{n2} \quad \cdots \quad a_{nn} \end{matrix} \right| + \left| \begin{matrix} a_{11} \quad a_{12} \quad \cdots \quad a_{1n} \ \vdots \quad \vdots \quad \cdots \quad \vdots \ b_1 \quad b_2 \quad \cdots \quad b_n \ \vdots \quad \vdots \quad \cdots \quad \vdots \ a_{n1} \quad a_{n2} \quad \cdots \quad a_{nn} \end{matrix} \right| $
推论: 若行列式某一行的元素都是m个元素的和. 则行列式可以写成m个行列式的和.
重点: 性质2, 性质3, 性质4.
练习1: 计算
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D=\left| \begin{matrix} 3 \quad 2 \quad 2 \quad 2 \\ 2 \quad 3 \quad 2 \quad 2 \\ 2\quad 2\quad 3 \quad 2 \\ 2 \quad 2 \quad 2 \quad 3\end{matrix} \right|
D=∣∣∣∣∣∣∣∣3222232222322223∣∣∣∣∣∣∣∣
分析: 各行元素之和为一定数, 利用性质4, 将2,3,4行全部加到第1行,然后利用性质3的推论, 将公因子提出来.
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D=\left| \begin{matrix} 9 \quad 9 \quad 9 \quad 9 \\ 2 \quad 3 \quad 2 \quad 2 \\ 2\quad 2\quad 3 \quad 2 \\ 2 \quad 2 \quad 2 \quad 3\end{matrix} \right| = 9\left| \begin{matrix} 1 \quad 1 \quad 1 \quad 1 \\ 2 \quad 3 \quad 2 \quad 2 \\ 2\quad 2\quad 3 \quad 2 \\ 2 \quad 2 \quad 2 \quad 3\end{matrix} \right|
D=∣∣∣∣∣∣∣∣9999232222322223∣∣∣∣∣∣∣∣=9∣∣∣∣∣∣∣∣1111232222322223∣∣∣∣∣∣∣∣
然后第一行乘上-2和后面的各行相加得到
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D=9\left| \begin{matrix} 1 \quad 1 \quad 1 \quad 1 \\ 0 \quad 1 \quad 0 \quad 0 \\ 0\quad0\quad 1 \quad 0 \\ 0 \quad 0 \quad 0 \quad 1\end{matrix} \right|=9
D=9∣∣∣∣∣∣∣∣1111010000100001∣∣∣∣∣∣∣∣=9
练习2: 设
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\alpha + \beta + \gamma=0, 求行列式\left| \begin{matrix} \alpha \quad \beta \quad \gamma \\ \gamma \quad \alpha \quad \beta \\ \beta \quad \gamma \quad \alpha \end{matrix} \right|的值
α+β+γ=0,求行列式∣∣∣∣∣∣αβγγαββγα∣∣∣∣∣∣的值
练习3:
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D=\left| \begin{matrix} 3 \quad 1 \quad -1 \quad 2 \\ -5 \quad 1 \quad 3 \quad -4 \\ 2\quad 0 \quad 1 \quad -1 \\ 1 \quad -5 \quad 3 \quad -3\end{matrix} \right|
D=∣∣∣∣∣∣∣∣31−12−513−4201−11−53−3∣∣∣∣∣∣∣∣
第二列有0元素,可以想办法把第二列元素尽可能变成0, 然后按照第二列展开.
根据性质4, 第一行乘-1, 加到第二行, 第一行乘5, 加到第4行
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∣
D=\left| \begin{matrix} 3 \quad 1 \quad -1 \quad 2 \\ -5 \quad 1 \quad 3 \quad -4 \\ 2\quad 0 \quad 1 \quad -1 \\ 1 \quad -5 \quad 3 \quad -3\end{matrix} \right| = \left| \begin{matrix} 3 \quad 1 \quad -1 \quad 2 \\ -8 \quad 0 \quad 4 \quad -6 \\ 2\quad 0 \quad 1 \quad -1 \\ 16 \quad 0 \quad -2 \quad -7\end{matrix} \right|
D=∣∣∣∣∣∣∣∣31−12−513−4201−11−53−3∣∣∣∣∣∣∣∣=∣∣∣∣∣∣∣∣31−12−804−6201−1160−2−7∣∣∣∣∣∣∣∣
=
−
∣
−
8
4
−
6
2
1
−
1
16
−
2
7
∣
=
−
∣
−
16
0
−
2
2
1
−
1
20
0
5
∣
=
−
∣
−
16
−
2
20
5
∣
=
40
=-\left| \begin{matrix} -8 \quad 4 \quad -6 \\ 2 \quad 1 \quad -1 \\ 16 \quad -2 \quad 7 \end{matrix}\right| = - \left| \begin{matrix} -16 \quad 0 \quad -2 \\ 2 \quad 1 \quad -1 \\ 20 \quad 0 \quad 5 \end{matrix}\right| = - \left| \begin{matrix} -16 \quad -2 \\ 20 \quad 5 \end{matrix}\right| = 40
=−∣∣∣∣∣∣−84−621−116−27∣∣∣∣∣∣=−∣∣∣∣∣∣−160−221−12005∣∣∣∣∣∣=−∣∣∣∣−16−2205∣∣∣∣=40
