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行列式(一)- 行列式介绍

行列式(一)- 行列式介绍

作者: mHubery | 来源:发表于2019-03-11 23:03 被阅读0次

    矩阵是可逆的,当且仅当它的行列式非零。

    考虑\boldsymbol{A} = \left[ a_i\!_j \right],a_i\!_j \neq 0\boldsymbol{A}的第二行和第三行都乘以a_1\!_1,然后再分别减去第一行适当的倍数,则\boldsymbol{A}行等价于下面两个矩阵:
    \begin{bmatrix} a_1\!_1 & a_1\!_2 & a_1\!_3 \\ a_1\!_1a_2\!_1 & a_1\!_1a_2\!_2 & a_1\!_1a_2\!_3 \\ a_1\!_1a_3\!_1 & a_1\!_1a_3\!_2 & a_1\!_1a_3\!_3 \end{bmatrix}\begin{bmatrix} a_1\!_1 & a_1\!_2 & a_1\!_3 \\ 0 & a_1\!_1a_2\!_2 - a_1\!_2a_2\!_1 & a_1\!_1a_2\!_3 - a_1\!_3a_2\!_1\\ 0 & a_1\!_1a_3\!_2 - a_1\!_2a_3\!_1& a_1\!_1a_3\!_3 - a_1\!_3a_3\!_1\end{bmatrix}
    由于\boldsymbol{A}可逆,故矩阵中(2,2)元素和(3,2)元素不同时为0.不妨假设(2,2)元素不等于零(否则,可以做一个行对换边变成这种情形)。在进行行化简:
    \boldsymbol{A}\begin{bmatrix} a_1\!_1 & a_1\!_2 & a_1\!_3 \\ 0 & a_1\!_1a_2\!_2 - a_1\!_2a_2\!_1 & a_1\!_1a_2\!_3 - a_1\!_3a_2\!_1\\ 0 & 0 & a_1\!_1\Delta\end{bmatrix}
    其中,\Delta=a_1\!_1a_2\!_2a_3\!_3 + a_1\!_2a_2\!_3a_3\!_1 + a_1\!_3a_2\!_1a_3\!_2 - a_1\!_1a_2\!_3a_3\!_2 - a_1\!_2a_2\!_1a_3\!_3 - a_1\!_3a_2\!_2a_3\!_1 \tag{1}
    由于\boldsymbol{A}可逆,故\Delta一定不等于零。我们称这个(1)式中的\Delta3 \times 3矩阵\boldsymbol{A}行列式

    2 \times 2矩阵\boldsymbol{A}的行列式:det \;\boldsymbol{A}= a_1\!_1a_2\!_2 - a_1\!_2a_2\!_11 \times 1矩阵\boldsymbol{A}的行列式:det \;\boldsymbol{A}=a_1\!_1。利用2 \times 2行列式来重写(1)中的行列式\Delta
    \begin{aligned}\Delta &= (a_1\!_1a_2\!_2a_3\!_3 - a_1\!_1a_2\!_3a_3\!_2) - (a_1\!_2a_2\!_1a_3\!_3 - a_1\!_2a_2\!_3a_3\!_1) + (a_1\!_3a_2\!_1a_3\!_2- a_1\!_3a_2\!_2a_3\!_1) \\ &= a_1\!_1 \times det \begin{bmatrix} a_2\!_2 & a_2\!_3 \\ a_3\!_2 & a_3\!_3\end{bmatrix} - a_1\!_2 \times det \begin{bmatrix} a_2\!_1 & a_2\!_3 \\ a_3\!_1 & a_3\!_3\end{bmatrix} + a_1\!_3 \times det \begin{bmatrix} a_2\!_1 & a_2\!_2 \\ a_3\!_1 & a_3\!_3\end{bmatrix}\end{aligned}
    为了简单,可写成\Delta=a_1\!_1 \times det \boldsymbol{A_1\!_1} - a_1\!_2 \times det \boldsymbol{A_1\!_2} + a_1\!_3 \times det \boldsymbol{A_1\!_3},其中\boldsymbol{A_1\!_1}, \boldsymbol{A_1\!_2}, \boldsymbol{A_1\!_3}\boldsymbol{A}中删除第一行和三列中之一列而得到。

    n \geq 2, n \times n矩阵\boldsymbol{A}=\left[ a_i\!_j \right]行列式是形如\pm a_1\!_j det\;\boldsymbol{A_1\!_j}n个项的和,其中加号和减号交替出现,元素a_1\!_1,\cdots,a_1\!_n来自于第一行,用符号表示为:
    \begin{aligned} det \;\boldsymbol{A} &= a_1\!_1 \times det \;\boldsymbol{A_1\!_1} - a_1\!_2 \times det \;\boldsymbol{A_1\!_2} + \cdots + (-1)^{1+n}a_1\!_n \times \boldsymbol{A_1\!_n } \\ &=\sum_{j=1}^{n}{(-1)^{1+j}det \;\boldsymbol{A_1\!_j}} \end{aligned}

    计算行列式det \;\boldsymbol{A},其中\boldsymbol{A}=\begin{bmatrix}1 & 5 & 0 \\ 2 & 4 & -1 \\ 0 & -2 & 0 \end{bmatrix}
    解:
    \begin{aligned} \Delta &= 1 \times det \begin{bmatrix}4 & -1 \\ -2 & 0\end{bmatrix} - 5 \times det \begin{bmatrix}2 & -1 \\ 0 & 0\end{bmatrix} + 0 \times det \begin{bmatrix}2 & 4 \\ 0 & -2\end{bmatrix} \\ &= 1(0 - 2) - 5(0-0) + 0(-4-0) = -2\end{aligned}
    方阵的行列式的另一个常用记号是利用一对竖线代替括号。这样,上式可写为:
    \Delta = 1 \times det \begin{vmatrix} 4 & -1 \\ -2 & 0\end{vmatrix} - 5 \times det \begin{vmatrix} 2 & -1 \\ 0 & 0\end{vmatrix} + 0 \times det \begin{vmatrix} 2 & 4 \\ 0 & -2\end{vmatrix}=\cdots=-2

    给定\boldsymbol{A}=\left[ a_i\!_j \right],\boldsymbol{A}的(i,j)余因子\boldsymbol{C_i\!_j}表示为:\boldsymbol{C_i\!_j}=(-1)^{i+j}detA_i\!_j,则det \;\boldsymbol{A}=a_1\!_1 \times \boldsymbol{C_1\!_1} + \cdots + a_1\!_n \times \boldsymbol{C_1\!_n}。这个公式称为按\boldsymbol{A}第一行的余因子展开式
    定理 1 \;n \times n矩阵的\boldsymbol{A}的行列式可按任意行或列的余因子展开式来计算,按第i行展开的余因子展开式为:det \;\boldsymbol{A} = a_1\!_1\boldsymbol{C_i\!_1} + \cdots + a_1\!_n\boldsymbol{C_1\!_n};按第j列的余因子展开式为:det \;\boldsymbol{A} = a_1\!_j\boldsymbol{C_1\!_j} + \cdots + a_n\!_j\boldsymbol{C_n\!_j}。(i,j)余因子中加号或减号取决于a_i\!_j在矩阵中的位置,而于a_i\!_j本身的符号无关。
    定理 2 \;\boldsymbol{A}为三角形,则det \boldsymbol{A}等于\boldsymbol{A}的主对角线上元素的乘积。

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