线性代数，临时抱佛脚速撸系列（1）行列式

2019-08-30 20:35

学习笔记  数学  线性代数 

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什么是行列式

$\begin{aligned}\left|\begin{array}{ll}{1} & {2} \\ {2} & {3}\end{array}\right| \end{aligned} \quad(\text { 二阶行列式 })$

$\left|\begin{array}{lll}{1} & {2} & {3} \\ {2} & {3} & {4} \\ {4} & {5} & {7}\end{array}\right| \quad(\text { 三阶行列式 })$

$\left|\begin{array}{cccc}{1} & {2} & {3} & {4} \\ {2} & {3} & {4} & {5} \\ {4} & {5} & {7} & {8} \\ {8} & {9} & {10} & {12}\end{array}\right| \quad(\text { 四阶行列式 })$

求二阶行列式

二阶行列式只需要把对角线的数值相乘然后相减，例如：

$\left|\begin{array}{ll}{1} & {2} \\ {2} & {3}\end{array}\right|=1 \times 3-2 \times 2=-1$

$\left|\begin{array}{ll}{3} & {4} \\ {5} & {6}\end{array}\right|=3 \times 6-4 \times 5=-2$

求多阶行列式

先将“左上开始的对角线”左上开始的对角线下面的数字都消成 0，例如：

$\left|\begin{array}{cccc}{1} & {2} & {3} & {4} \\ {0} & {-1} & {-2} & {-3} \\ {0} & {0} & {1} & {1} \\ {0} & {0} & {0} & {1}\end{array}\right|$

然后将“左上开始的对角线”的值相乘，得到结果。

计算过程中，有以下性质可以使用：

• 某行（列）加上或减去另一行（列）的几倍，行列式不变
  例如：

  $\left|\begin{array}{lll}{1} & {2} & {3} \\ {2} & {3} & {4} \\ {4} & {5} & {7}\end{array}\right| \stackrel{ \mathbf{r}_{2} -2 \mathbf{r}_{1} }{=} \left|\begin{array}{ccc}{1} & {2} & {3} \\ {2-2 \times 1} & {3-2 \times 2} & {4-2 \times 3} \\ {4} & {5} & {7}\end{array}\right|=\left|\begin{array}{ccc}{1} & {2} & {3} \\ {0} & {-1} & {-2} \\ {4} & {5} & {7}\end{array}\right|$

• 某行（列）乘 k，等于 k 乘此行列式
  比如有以下行列式等于-1：

  $\left|\begin{array}{cccc}{1} & {2} & {3} & {4} \\ {2} & {3} & {4} & {5} \\ {4} & {5} & {7} & {8} \\ {8} & {9} & {10} & {12}\end{array}\right|=-1$

  求下面这个行列式

  $\left|\begin{array}{cccc}{2} & {4} & {6} & {8} \\ {2} & {3} & {4} & {5} \\ {12} & {15} & {21} & {24} \\ {8} & {9} & {10} & {12}\end{array}\right|$

  根据该性质，我们可以得到：

  $\left|\begin{array}{cccc}{2} & {4} & {6} & {8} \\ {2} & {3} & {4} & {5} \\ {12} & {15} & {21} & {24} \\ {8} & {9} & {10} & {12}\end{array}\right|=2 \times 3 \times\left|\begin{array}{cccc}{1} & {2} & {3} & {4} \\ {2} & {3} & {4} & {5} \\ {4} & {5} & {7} & {8} \\ {8} & {9} & {10} & {12}\end{array}\right|=2 \times 3 \times(-1)=-6$

• 互换两行（列），行列式变号
  例如，已知下面这个行列式：

  $\left|\begin{array}{cccc}{1} & {2} & {3} & {4} \\ {2} & {3} & {4} & {5} \\ {4} & {5} & {7} & {8} \\ {8} & {9} & {10} & {12}\end{array}\right|=-1$

  求：

  $\left|\begin{array}{cccc}{2} & {3} & {4} & {5} \\ {1} & {2} & {3} & {4} \\ {4} & {5} & {7} & {8} \\ {8} & {9} & {10} & {12}\end{array}\right|$

  可以看到这两个行列式第一行和第二行互换，第三行和第四行相同。所以：

  $\left|\begin{array}{llll}{2} & {3} & {4} & {5} \\ {1} & {2} & {3} & {4} \\ {4} & {5} & {7} & {8} \\ {8} & {9} & {10} & {12}\end{array}\right| \stackrel{ \mathbf{r}_{1} \leftrightarrow \mathbf{r}_{2} }{=} -1 \times\left|\begin{array}{llll}{1} & {2} & {3} & {4} \\ {2} & {3} & {4} & {5} \\ {4} & {5} & {7} & {8} \\ {8} & {9} & {10} & {12}\end{array}\right|=-1 \times(-1)=1$

计算过程可以省略掉一些步骤，直接把过程写到等号下：

行列式的一些公式

1

对于一个 n 行 n 列的行列式：

$\left|\begin{array}{cccc}{x} & {a} & {\cdots} & {a} \\ {a} & {x} & {\cdots} & {a} \\ {\vdots} & {\vdots} & {\ddots} & {\vdots} \\ {a} & {a} & {\cdots} & {x}\end{array}\right|=(x-a)^{n-1}[x+(n-1) a]$

例如，这个行列式：

$\left|\begin{array}{llll}{2} & {3} & {3} & {3} \\ {3} & {2} & {3} & {3} \\ {3} & {3} & {2} & {3} \\ {3} & {3} & {3} & {2}\end{array}\right|$

其中 $x=2$, $a=3$, $n=4$，所以根据公式：

$\left|\begin{array}{cccc}{2} & {3} & {3} & {3} \\ {3} & {2} & {3} & {3} \\ {3} & {3} & {2} & {3} \\ {3} & {3} & {3} & {2}\end{array}\right|=(2-3)^{4-1}[2+(4-1) \times 3]=-11$

2

$\left|\begin{array}{cccc}{1} & {1} & {\cdots} & {1} \\ {x_{1}} & {x_{2}} & {\cdots} & {x_{n}} \\ {x_{1}^{2}} & {x_{2}^{2}} & {\cdots} & {x_{n}^{2}} \\ {\vdots} & {\vdots} & {\ddots} & {\vdots} \\ {x_{1}^{n-1}} & {x_{2}^{n-1}} & {\cdots} & {x_{n}^{n-1}}\end{array}\right| = \begin{array}{l}{\left(x_{n}-x_{n-1}\right)\left(x_{n}-x_{n-2}\right)\left(x_{n}-x_{n-3}\right) \cdots \cdots\left(x_{n}-x_{1}\right)} \\ {\times\left(x_{n-1}-x_{n-2}\right)\left(x_{n-1}-x_{n-3}\right) \cdots \cdots\left(x_{n-1}-x_{1}\right)} \\ {\times \cdots \cdots} \\ {\times\left(x_{2}-x_{1}\right)}\end{array}$

例如：

$\left|\begin{array}{cccc}{1} & {1} & {1} & {1} \\ {3} & {4} & {5} & {6} \\ {3^{2}} & {4^{2}} & {5^{2}} & {6^{2}} \\ {3^{3}} & {4^{3}} & {5^{3}} & {6^{3}}\end{array}\right|$

$\mathrm{x}_{1}=3 \quad \mathrm{x}_{2}=4 \quad \mathrm{x}_{3}=5 \quad \mathrm{x}_{4}=6 \quad \mathrm{n}=4$

$\left|\begin{array}{cccc}{1} & {1} & {1} & {1} \\ {3} & {4} & {5} & {6} \\ {3^{2}} & {4^{2}} & {5^{2}} & {6^{2}} \\ {3^{3}} & {4^{3}} & {5^{3}} & {6^{3}}\end{array}\right|=(6-5)(6-4)(6-3)(5-4)(5-3)(4-3)=12$

3

1. 两行（列）相同或者成比例时，行列式为0
2. 某行（列）为两项相加减时，行列式可拆成两个行列式相加减

例题：

$\text { 已知 } \left|\begin{array}{lll}{a_{1}} & {b_{1}} & {c_{1}} \\ {a_{2}} & {b_{2}} & {c_{2}} \\ {a_{3}} & {b_{3}} & {c_{3}}\end{array}\right|=1, \quad \text { 试求 } \left|\begin{array}{lll}{a_{1}+c_{1}} & {b_{1}} & {a_{1}+b_{1}} \\ {a_{2}+c_{2}} & {b_{2}} & {a_{2}+b_{2}} \\ {a_{3}+c_{3}} & {b_{3}} & {a_{3}+b_{3}}\end{array}\right|$

解题过程：

$\left|\begin{array}{lll}{a_{1}+c_{1}} & {b_{1}} & {a_{1}+b_{1}} \\ {a_{2}+c_{2}} & {b_{2}} & {a_{2}+b_{2}} \\ {a_{3}+c_{3}} & {b_{3}} & {a_{3}+b_{3}}\end{array}\right|=\left|\begin{array}{ccc}{a_{1}} & {b_{1}} & {a_{1}+b_{1}} \\ {a_{2}} & {b_{2}} & {a_{2}+b_{2}} \\ {a_{3}} & {b_{3}} & {a_{3}+b_{3}}\end{array}\right|+\left|\begin{array}{ccc}{c_{1}} & {b_{1}} & {a_{1}+b_{1}} \\ {c_{2}} & {b_{2}} & {a_{2}+b_{2}} \\ {c_{3}} & {b_{3}} & {a_{3}+b_{3}}\end{array}\right|$

$=\left|\begin{array}{lll}{a_{1}} & {b_{1}} & {a_{1}} \\ {a_{2}} & {b_{2}} & {a_{2}} \\ {a_{3}} & {b_{3}} & {a_{3}}\end{array}\right|+\left|\begin{array}{lll}{a_{1}} & {b_{1}} & {b_{1}} \\ {a_{2}} & {b_{2}} & {b_{2}} \\ {a_{3}} & {b_{3}} & {b_{3}}\end{array}\right|+\left|\begin{array}{lll}{c_{1}} & {b_{1}} & {a_{1}} \\ {c_{2}} & {b_{2}} & {a_{2}} \\ {c_{3}} & {b_{3}} & {a_{3}}\end{array}\right|+\left|\begin{array}{ccc}{c_{1}} & {b_{1}} & {b_{1}} \\ {c_{2}} & {b_{2}} & {b_{2}} \\ {c_{3}} & {b_{3}} & {b_{3}}\end{array}\right|$

$=0+0+\left|\begin{array}{ccc}{c_{1}} & {b_{1}} & {a_{1}} \\ {c_{2}} & {b_{2}} & {a_{2}} \\ {c_{3}} & {b_{3}} & {a_{3}}\end{array}\right|+0=-\left|\begin{array}{ccc}{a_{1}} & {b_{1}} & {c_{1}} \\ {a_{2}} & {b_{2}} & {c_{2}} \\ {a_{3}} & {b_{3}} & {c_{3}}\end{array}\right|=-1$

4 求余子式，代数余子式

余子式用 $M$ 表示，代数余子式用 $A$ 表示。
比如，求 $\left|\begin{array}{ccc}{1} & {2} & {3} \\ {5} & {6} & {7} \\ {9} & {10} & {11}\end{array}\right|$ 中 $a_{23}$ 的余子式，就是求 $M_{23}$，求 $a_{23}$ 的代数余子式，就是求 $A_{23}$。

求余子式 $M_{23}$，消掉第二行和第三列，生成一个新行列式，求新行列式的值：

$\begin{array}{l}{M_{23}=\left|\begin{array}{cc}{1} & {2} \\ {9} & {10}\end{array}\right|=-8} \\ {M_{12}=\left|\begin{array}{cc}{5} & {7} \\ {9} & {11}\end{array}\right|=-8}\end{array}$

求代数余子式 $A_{23}$，即是余子式 $M_{23}$ 的值再乘于 -1 的行加列次方（$(-1)^{2+3}$
）：

$\begin{array}{l}{A_{23}=(-1)^{2+3} \cdot M_{23}=-1 \times(-8)=8} \\ {A_{12}=(-1)^{1+2} \cdot M_{12}=-1 \times(-8)=8}\end{array}$

5

$\begin{array}{l}{D=a_{i 1} A_{i 1}+a_{i 2} A_{i 2}+\cdots \cdots+a_{i n} A_{i n}} \quad\text{(第i行)} \\ {D=a_{1 j} A_{1 j}+a_{2 j} A_{2 j}+\cdots \cdots+a_{n j} A_{n j}}\quad\text{(第j行)} \end{array}$

例如，取第一行套用上面的行公式：

$\left|\begin{array}{ccc}{1} & {2} & {3} \\ {5} & {6} & {7} \\ {9} & {10} & {11}\end{array}\right|=a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13}$

$=a_{11}(-1)^{1+1} M_{11}+a_{12}(-1)^{1+2} M_{12}+a_{13}(-1)^{1+3} M_{13}$

$=1 \times(-1)^{1+1} \times\left|\begin{array}{cc}{6} & {7} \\ {10} & {11}\end{array}\right|+2 \times(-1)^{1+2} \times\left|\begin{array}{cc}{5} & {7} \\ {9} & {11}\end{array}\right|+3 \times(-1)^{1+3} \times\left|\begin{array}{cc}{5} & {6} \\ {9} & {10}\end{array}\right|$

套用上面的列公式：

$\left|\begin{array}{ccc}{1} & {2} & {3} \\ {5} & {6} & {7} \\ {9} & {10} & {11}\end{array}\right|=a_{11} A_{11}+a_{21} A_{21}+a_{31} A_{31}$

$=a_{11}(-1)^{1+1} M_{11}+a_{21}(-1)^{2+1} M_{21}+a_{31}(-1)^{3+1} M_{31}$

$=1 \times(-1)^{1+1} \times\left|\begin{array}{cc}{6} & {7} \\ {10} & {11}\end{array}\right|+5 \times(-1)^{2+1} \times\left|\begin{array}{cc}{2} & {3} \\ {10} & {11}\end{array}\right|+9 \times(-1)^{3+1} \times\left|\begin{array}{cc}{2} & {3} \\ {6} & {7}\end{array}\right|$

特殊情况，如果有某一行或者某一列有 0，直接用有 0 的那一行来计算：

$\left|\begin{array}{lll}{1} & {2} & {3} \\ {4} & {0} & {5} \\ {6} & {0} & {7}\end{array}\right|=2 \times(-1)^{1+2} \times\left|\begin{array}{cc}{4} & {5} \\ {6} & {7}\end{array}\right|+0 \times(-1)^{2+2} \times\left|\begin{array}{cc}{1} & {3} \\ {6} & {7}\end{array}\right|+0 \times(-1)^{3+2} \times\left|\begin{array}{cc}{1} & {3} \\ {4} & {5}\end{array}\right|$

$=2 \times(-1)^{1+2} \times\left|\begin{array}{ll}{4} & {5} \\ {6} & {7}\end{array}\right|$

6 多个 $A$ 或 $M$ 相加减

例，已知 $D=\left|\begin{array}{cccc}{1} & {2} & {3} & {4} \\ {5} & {6} & {7} & {8} \\ {9} & {10} & {11} & {12} \\ {13} & {14} & {15} & {16}\end{array}\right|$， 求下面 3 个问题：

1. $3 A_{11}+4 A_{12}+5 A_{13}+6 A_{14}$
2. $3 A_{11}+4 A_{21}+5 A_{31}+6 A_{41}$
3. $3 M_{11}+4 M_{21}+5 M_{31}+6 M_{41}$

求所有 $A$ 相加，说先找出所有 $A$ 对应的位置，然后用 $A$ 前面的系数替换这些数，得到一个新的行列式，求新的行列式的值，即是解：

$3 \mathrm{A}_{11}+4 \mathrm{A}_{12}+5 \mathrm{A}_{13}+6 \mathrm{A}_{14}=\left|\begin{array}{cccc}{3} & {4} & {5} & {6} \\ {5} & {6} & {7} & {8} \\ {9} & {10} & {11} & {12} \\ {13} & {14} & {15} & {16}\end{array}\right|$

$\begin{array}{llll}{ 3 \mathrm{A}_{11}+4 \mathrm{A}_{21}+5 \mathrm{A}_{31}+6 \mathrm{A}_{41}=} & {\left|\begin{array}{llll}{3} & {2} & {3} & {4} \\ {4} & {6} & {7} & {8} \\ {6} & {14} & {15} & {16}\end{array}\right|}\end{array}$

求所有 $M$ 相加，首先根据 $M$ 和 $A$ 的关系，可以将 $M$ 转化为 $A$，然后按照 $A$ 的求法求解即可：

$\begin{array}{l}{A_{11}=(-1)^{1+1} \cdot M_{11}=M_{11} \quad\rightarrow\quad M_{11}=A_{11}} \\ {A_{21}=(-1)^{2+1} \cdot M_{21}=-M_{21} \quad\rightarrow\quad M_{21}=-A_{21}} \\ {A_{31}=(-1)^{3+1} \cdot M_{31}=M_{31} \quad\rightarrow\quad M_{31}=A_{31}} \\ {A_{41}=(-1)^{4+1} \cdot M_{41}=-M_{41} \quad\rightarrow\quad M_{41}=-A_{41}}\end{array}$

$\begin{aligned} & 3 \mathrm{M}_{11}+4 \mathrm{M}_{21}+5 \mathrm{M}_{31}+6 \mathrm{M}_{41} = 3 \mathrm{A}_{11}-4 \mathrm{A}_{23}+5 \mathrm{A}_{33}-6 \mathrm{A}_{43} \end{aligned}$

7 给一方程组，判断其解的情况

例如，判断下列方程组是否有唯一解：

$\left\{\begin{aligned} x_{1}+2 x_{2}+3 x_{3} &=0 \\ 4 x_{1}+5 x_{2}+6 x_{3} &=0 \\ 7 x_{1}+8 x_{2}+9 x_{3} &=0 \end{aligned}\right.$

把系数提取成行列式：

$D=\left|\begin{array}{lll}{1} & {2} & {3} \\ {4} & {5} & {6} \\ {7} & {8} & {9}\end{array}\right|$

算行列式的值。根据行列式的值，对照上表即可。

提取行列式时，如果方程组有系数为 0 的未知数，需要补全：

$\left\{\begin{array}{l}{x_{1}+3 x_{3}=0} \\ {x_{2}+4 x_{3}=0} \\ {x_{1}+5 x_{2}=0}\end{array} \Longrightarrow\left\{\begin{array}{l}{x_{1}+0 x_{2}+3 x_{3}=0} \\ {0 x_{1}+x_{2}+4 x_{3}=0} \\ {x_{1}+5 x_{2}+0 x_{3}=0}\end{array}\right.\right.$

即是提取出：

$D=\left|\begin{array}{lll}{1} & {0} & {3} \\ {0} & {1} & {4} \\ {1} & {5} & {0}\end{array}\right|$