Section03_行列式的计算性质

Q 如何计算行列式

1. $D \Rightarrow \left\vert \begin{matrix} a_{11} & & & \huge0 \\ & a_{22} & & \\ & & \ddots & \\ \huge * & & & a_{nn} \\ \end{matrix} \right\vert$ 或 $\left\vert \begin{matrix} a_{11} & & & \huge * \\ & a_{22} & & \\ & & \ddots & \\ \huge 0 & & & a_{nn} \\ \end{matrix} \right\vert$
2. 降阶

化为三角矩阵

1. $D = D^{\intercal}$
2. 对调两行或列，行列式变为相反数
3. 一行(或一列)有公因子，可提取
4. $\left\vert \begin{matrix} a_{1} + b_{1} & c_{1} \\ a_{2} + b_{2} & c_{2} \end{matrix} \right\vert = \left\vert \begin{matrix} a_{1} & c_{1} \\ a_{2} & c_{2} \\ \end{matrix} \right\vert + \left\vert \begin{matrix} b_{1} & c_{1} \\ b_{2} & c_{2} \\ \end{matrix} \right\vert$
5. 一行的 $k$ 倍加到另一行 (一列的 $k$ 倍加到另一列)，行列式不变

降阶性质

1. 某一行或某一列的元素及其代数余子式之积的和等于该行列式
   • $\displaystyle \vert A \vert =\sum_{i=1}^{n}a_{ij}A_{ij}$
   • $\displaystyle \vert A \vert = \sum_{j=1}^{n}a_{ij}A_{ij}$
2. $\displaystyle 0 = \sum_{i=1}^{n}a_{ij_{1}}A_{ij_{2}}\quad(j_{1}\ne j_{2})$ (某一行的元素与其他行相对应的代数余子式之积的和为$0$)

例题

1. $A = \left( \begin{matrix} 1 & 2 & -1 \\ 2 & 3 & 1 \\ 1 & 4 & -5 \end{matrix} \right)$，求 $\det(A)$ $$\begin{array}{ll} &\det(A) = \left\vert \begin{matrix} 1 & 2 & -1 \\ 2 & 3 & 1 \\ 1 & 4 & -5 \end{matrix} \right\vert = \left\vert \begin{matrix} 1 & 2 & -1 \\ 0 & -1 & 3 \\ 0 & 0 & 2 \end{matrix} \right\vert \end{array} = -2$$
2. $A = \left( \begin{matrix} 3 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 3 \end{matrix} \right)$，求 $\det(A)$ $$\det(A) = \left\vert \begin{matrix} 5 & 5 & 5 \\ 1 & 3 & 1 \\ 1 & 1 & 3 \end{matrix} \right\vert = 5 \left\vert \begin{matrix} 1 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix} \right\vert = 20$$
3. $\left\vert \begin{matrix} a^{2} & 2a & 0 & 0 \\ 1 & a^{2} & 2a & 0 \\ 0 & 1 & a^{2} & 2a \\ 0 & 0 & 1 & a^{2} \\ \end{matrix} \right\vert$ $$\begin{array}{ll} & \left\vert \begin{matrix} a^{2} & 2a & 0 & 0 \\ 1 & a^{2} & 2a & 0 \\ 0 & 1 & a^{2} & 2a \\ 0 & 0 & 1 & a^{2} \\ \end{matrix} \right\vert = a^{2} \left\vert \begin{matrix} a^{2} & 2a & 0 \\ 1 & a^{2} & 2a \\ 0 & 1 &a^{2} \end{matrix} \right\vert - 2a \left\vert \begin{matrix} 1 & 2a & 0 \\ 0 & a^{2} & 2a \\ 0 & 1 & a^{2} \end{matrix} \right\vert \\ & = a^{2}a^{2}(a^{4} - 2a) - a^{2}2aa^{2} - 2a (a^{4}-2a) \\ & = a^{8} - 2a^{5} -2a^{5} -2a^{5} + 4a^{2} \\ & = a^{8} - 6a^{5} + 4a^{2} \end{array}$$
4. $D = \left\vert \begin{matrix} 1 & a & 0 & 0 \\ 0 & 1 & a & 0 \\ 0 & 0 & 1 & a \\ a & 0 & 0 & 1 \\ \end{matrix} \right\vert$ $$D=1 - a \left\vert \begin{matrix} 0 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \end{matrix} \right\vert = 1 + a^{2}(-a^{2}) = 1-a^{4}$$