3.3. Laplace 展开定理

我们在讨论面积元和体积元的时候, 观察到 3 阶行列式可以化为 2 阶行列式的展开: $$\left|\begin{array}{ccc}{u}_1& u_2& u_3\\ v_1& v_2& v_3\\ w_1& w_2& w_3\end{array}\right|=u_1\left|\begin{array}{cc}{v}_2& v_3\\ w_2& w_3\end{array}\right|-u_2\left|\begin{array}{cc}{v}_1& v_3\\ w_1& w_3\end{array}\right|+u_3\left|\begin{array}{cc}{v}_1& v_2\\ w_1& w_2\end{array}\right|$$实际上 $n$ 阶行列式也可以化为低阶行列式的展开, 这个称为 Laplace 展开定理.

余子式与展开定理

给定 $n$ 阶方程$$A=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ \cdots & \cdots & \cdots & \cdots \\ a_{n1}& a_{n2}& \cdots & a_{nn}\end{array}\right]$$

定义 3.3.1. $A$ 的第 $i$ 行第 $j$ 列余子式 $A_{ij}$ 定义为把 $A$ 的第 $i$ 行和第 $j$ 列去掉后得到的 $n-1$ 阶方阵的行列式, 即

$$A_{ij}=\left|\begin{array}{ccccc}{a}_{11}& \cdots & \cancel{a_{1j}}& \cdots & a_{1n}\\ \cdots & \cdots & ⋮& \cdots & \cdots \\ \cancel{a_{i1}}& \cdots & \cancel{a_{ij}}& \cdots & \cancel{a_{in}}\\ \cdots & \cdots & ⋮& \cdots & \cdots \\ a_{n1}& \cdots & \cancel{a_{nj}}& \cdots & a_{nn}\end{array}\right|$$

例子 3.3.2. $A=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]$, 则$$A_{12}=\left|\begin{array}{cc}4& 6\\ 7& 9\end{array}\right|=-6A_{23}=\left|\begin{array}{cc}1& 2\\ 7& 8\end{array}\right|=-6$$

例子 3.3.3. 体积元的展开公式$$\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|=a_{11}\left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|-a_{12}\left|\begin{array}{cc}{a}_{21}& a_{23}\\ a_{31}& a_{33}\end{array}\right|+a_{13}\left|\begin{array}{cc}{a}_{21}& a_{22}\\ a_{31}& a_{32}\end{array}\right|$$可以写成$$\det A=a_{11}{A}_{11}-a_{12}{A}_{12}+a_{13}{A}_{13}$$

命题 3.3.4. 设 $A=(a_{ij})$ 是 $n$ 阶方阵, 则按照第一行展开有如下公式成立$$\det A=\sum _{j=1}^n(-1{)}^{1+j}{a}_{1j}{A}_{1j}$$

证明: 记 $A$ 的行向量为$$\begin{array}{rl}{\vec{\alpha }}_1& =(a_{11},a_{12},\cdots ,a_{1n})=a_{11}{\vec{e}}_1+a_{12}{\vec{e}}_2+\cdots +a_{1n}{\vec{e}}_n\\ {\vec{\alpha }}_2& =(a_{21},a_{22},\cdots ,a_{2n})=a_{21}{\vec{e}}_1+a_{22}{\vec{e}}_2+\cdots +a_{2n}{\vec{e}}_n\\ & ⋮\\ {\vec{\alpha }}_n& =(a_{n1},a_{n2},\cdots ,a_{nn})=a_{n1}{\vec{e}}_1+a_{n2}{\vec{e}}_2+\cdots +a_{nn}{\vec{e}}_n\end{array}$$

由行列式的外积定义, 我们有$${\vec{\alpha }}_1\wedge {\vec{\alpha }}_2\wedge \cdots \wedge {\vec{\alpha }}_n=(\det A){\vec{e}}_1\wedge {\vec{e}}_2\wedge \cdots \wedge {\vec{e}}_n$$另一方面, 代入 ${\vec{\alpha }}_1$ 的表达式$$\begin{array}{r}{\vec{\alpha }}_1\wedge {\vec{\alpha }}_2\wedge \cdots \wedge {\vec{\alpha }}_n=(a_{11}{\vec{e}}_1+a_{12}{\vec{e}}_2+\cdots +a_{1n}{\vec{e}}_n)\wedge {\vec{\alpha }}_2\wedge \cdots \wedge {\vec{\alpha }}_n\end{array}$$

考虑其中的第一项 $a_{11}{\vec{e}}_1\wedge {\vec{\alpha }}_2\wedge \cdots \wedge {\vec{\alpha }}_n$, 记$$\begin{array}{rl}{\vec{\alpha }}_2& =a_{21}{\vec{e}}_1+a_{22}{\vec{e}}_2+\cdots +a_{2n}{\vec{e}}_n=a_{21}{\vec{e}}_1+{\vec{\beta }}_2\\ & ⋮\\ {\vec{\alpha }}_n& =a_{n1}{\vec{e}}_1+a_{n2}{\vec{e}}_2+\cdots +a_{nn}{\vec{e}}_n=a_{n1}{\vec{e}}_1+{\vec{\beta }}_n\end{array}$$则行向量 ${\vec{\beta }}_2,\cdots ,{\vec{\beta }}_n$ 组成把 $A$ 的第 1 行第 1 列去掉后得到的 $n-1$ 阶方阵. 由外积代数我们有$${\vec{\beta }}_2\wedge {\vec{\beta }}_3\wedge \cdots \wedge {\vec{\beta }}_n=A_{11}{\vec{e}}_2\wedge {\vec{e}}_3\wedge \cdots \wedge {\vec{e}}_n$$

另一方面, 由外积代数关系 ${\vec{e}}_1\wedge {\vec{e}}_1=0$,$${\vec{e}}_1\wedge {\vec{\alpha }}_2={\vec{e}}_1\wedge (a_{21}{\vec{e}}_1+{\vec{\beta }}_2)={\vec{e}}_1\wedge {\vec{\beta }}_2$$类似的我们得到$$\begin{array}{rl}& {\vec{e}}_1\wedge {\vec{\alpha }}_2\wedge \cdots \wedge {\vec{\alpha }}_n\\ =& {\vec{e}}_1\wedge (a_{21}{\vec{e}}_1+{\vec{\beta }}_2)\wedge \cdots \wedge (a_{n1}{\vec{e}}_1+{\vec{\beta }}_n)\\ =& {\vec{e}}_1\wedge {\vec{\beta }}_2\wedge \cdots \wedge {\vec{\beta }}_n\end{array}$$

因此 $\det A$ 展开的第一项可以写成$$\begin{array}{r}{a}_{11}{\vec{e}}_1\wedge {\vec{\alpha }}_2\wedge \cdots \wedge {\vec{\alpha }}_n=a_{11}{\vec{e}}_1\wedge {\vec{\beta }}_2\wedge \cdots \wedge {\vec{\beta }}_n=a_{11}{A}_{11}{\vec{e}}_1\wedge {\vec{e}}_2\wedge \cdots \wedge {\vec{e}}_n\end{array}$$

类似的, $\det A$ 展开的第 $j$ 项可以计算为$$\begin{array}{r}{a}_{1j}{\vec{e}}_j\wedge {\vec{\alpha }}_2\wedge \cdots \wedge {\vec{\alpha }}_n=a_{1j}{A}_{1j}{\vec{e}}_j\wedge {\vec{e}}_1\wedge \cdots \cancel{{\vec{e}}_j}\cdots \wedge {\vec{e}}_n\end{array}$$我们可以用外积的反对称把 ${\vec{e}}_j$ 从最前面移动到第 $j$ 个位置, 共需要交换 $(j-1)$ 次. 所以$$\begin{array}{rl}& a_{1j}{A}_{1j}{\vec{e}}_j\wedge {\vec{e}}_1\wedge \cdots \cancel{{\vec{e}}_j}\cdots \wedge {\vec{e}}_n\\ =& (-1{)}^{1+j}{a}_{1j}{A}_{1j}{\vec{e}}_1\wedge \cdots {\vec{e}}_j\cdots \wedge {\vec{e}}_n\end{array}$$

把所有项加起来, 我们得到$$\begin{array}{rl}& {\vec{\alpha }}_1\wedge {\vec{\alpha }}_2\wedge \cdots \wedge {\vec{\alpha }}_n\\ =& (a_{11}{\vec{e}}_1+a_{12}{\vec{e}}_2+\cdots +a_{1n}{\vec{e}}_n)\wedge {\vec{\alpha }}_2\wedge \cdots \wedge {\vec{\alpha }}_n\\ =& \left(\sum _{j=1}^n(-1{)}^{1+j}{a}_{1j}{A}_{1j}\right){\vec{e}}_1\wedge {\vec{e}}_2\wedge \cdots \wedge {\vec{e}}_n\end{array}$$比较行列式的外积定义, 我们证明了命题所述$$\det A=\sum _{j=1}^n(-1{)}^{1+j}{a}_{1j}{A}_{1j}$$

$\square$

命题 3.3.5. 设 $A=(a_{ij})$ 是 $n$ 阶方阵, 则按照第一列展开有如下公式成立$$\det A=\sum _{i=1}^n(-1{)}^{i+1}{a}_{i1}{A}_{i1}$$

证明: 记 $A$ 的行向量为$$\begin{array}{rl}{\vec{\alpha }}_1& =a_{11}{\vec{e}}_1+a_{12}{\vec{e}}_2+\cdots +a_{1n}{\vec{e}}_n=a_{11}{\vec{e}}_1+{\vec{\beta }}_1\\ {\vec{\alpha }}_2& =a_{21}{\vec{e}}_1+a_{22}{\vec{e}}_2+\cdots +a_{2n}{\vec{e}}_n=a_{21}{\vec{e}}_1+{\vec{\beta }}_2\\ & ⋮\\ {\vec{\alpha }}_n& =a_{n1}{\vec{e}}_1+a_{n2}{\vec{e}}_2+\cdots +a_{nn}{\vec{e}}_n=a_{n1}{\vec{e}}_1+{\vec{\beta }}_n\end{array}$$

利用外积的性质, 我们可以展开得到$$\begin{array}{rl}{\vec{\alpha }}_1\wedge {\vec{\alpha }}_2\wedge \cdots \wedge {\vec{\alpha }}_n=& (a_{11}{\vec{e}}_1+{\vec{\beta }}_1)\wedge (a_{21}{\vec{e}}_1+{\vec{\beta }}_2)\wedge \cdots \wedge (a_{n1}{\vec{e}}_1+{\vec{\beta }}_n)\\ =& \sum _{i=1}^n(-1{)}^{i+1}{a}_{i1}{\vec{e}}_1\wedge {\vec{\beta }}_1\wedge \cdots \cancel{{\vec{\beta }}_i}\cdots \wedge {\vec{\beta }}_n\\ =& \sum _{i=1}^n(-1{)}^{i+1}{a}_{i1}{A}_{i1}{\vec{e}}_1\wedge {\vec{e}}_2\wedge \cdots \wedge {\vec{e}}_n\end{array}$$这里最后一步用了和前述命题证明中类似的计算.

$\square$

定理 3.3.6 (Laplace 展开定理). 设 $A=(a_{ij})$ 为 $n$ 阶方阵, $A_{ij}$ 是第 $i$ 行 $j$ 列的余子式.

•	行展开公式: 对任意第 $i$ 行$$\det A=\sum _{j=1}^n(-1{)}^{i+j}{a}_{ij}{A}_{ij}$$
•	列展开公式: 对任意第 $j$ 列$$\det A=\sum _{i=1}^n(-1{)}^{i+j}{a}_{ij}{A}_{ij}$$

证明: 我们在前述命题中证明了第 $1$ 行的展开公式和第 $1$ 列的展开公式. 一般的情况证明方法类似, 证明细节留作练习.

$\square$

例子 3.3.7.

按照第 $1$ 行展开计算$$\begin{array}{rl}\left|\begin{array}{ccc}1& 2& 3\\ 0& -1& 4\\ 2& 1& -2\end{array}\right|=& 1\cdot \left|\begin{array}{cc}-1& 4\\ 1& -2\end{array}\right|-2\cdot \left|\begin{array}{cc}0& 4\\ 2& -2\end{array}\right|+3\cdot \left|\begin{array}{cc}0& -1\\ 2& 1\end{array}\right|\\ =& 1\cdot (2-4)-2\cdot (0-8)+3\cdot (0-(-2))=20\end{array}$$

按照第 $2$ 列展开计算$$\begin{array}{rl}\left|\begin{array}{ccc}1& 2& 3\\ 0& -1& 4\\ 2& 1& -2\end{array}\right|=& -2\cdot \left|\begin{array}{cc}0& 4\\ 2& -2\end{array}\right|+(-1)\cdot \left|\begin{array}{cc}1& 3\\ 2& -2\end{array}\right|-1\cdot \left|\begin{array}{cc}1& 3\\ 0& 4\end{array}\right|\\ =& (-2)\cdot (0-8)-1\cdot (-2-6)-1\cdot (4-0)=20\end{array}$$

例子 3.3.8. 我们知道对于上三角方阵$$\left|\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& \cdots & a_{1n}\\ 0& a_{22}& a_{23}& \cdots & a_{2n}\\ 0& 0& a_{33}& \cdots & a_{3n}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \cdots & a_{nn}\end{array}\right|=a_{11}{a}_{22}\cdots a_{nn}$$这个结论可以很容易地通过展开定理理解. 我们把行列式按照第 1 列展开得到$$\left|\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& \cdots & a_{1n}\\ 0& a_{22}& a_{23}& \cdots & a_{2n}\\ 0& 0& a_{33}& \cdots & a_{3n}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \cdots & a_{nn}\end{array}\right|=a_{11}\left|\begin{array}{cccc}{a}_{22}& a_{23}& \cdots & a_{2n}\\ 0& a_{33}& \cdots & a_{3n}\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & a_{nn}\end{array}\right|$$重复这个过程就得到行列式为 $a_{11}{a}_{22}\cdots a_{nn}$.

转置与行列对称性

定义 3.3.9. 给定 $m\times n$ 矩阵 $A$, 将 $A$ 的行排成列列排成行, 得到的 $n\times m$ 矩阵称为 $A$ 的转置, 记为 $A^T$$$A=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ \cdots & \cdots & \cdots & \cdots \\ a_{m1}& a_{m2}& \cdots & a_{mn}\end{array}\right]A^T=\left[\begin{array}{cccc}{a}_{11}& a_{21}& \cdots & a_{m1}\\ a_{12}& a_{22}& \cdots & a_{m2}\\ \cdots & \cdots & \cdots & \cdots \\ a_{1n}& a_{2n}& \cdots & a_{mn}\end{array}\right]$$

可以看出, 如果 $A$ 是 $n$ 阶方阵, 则 $A^T$ 也是 $n$ 阶方阵.

命题 3.3.10. 设 $A$ 是 $m\times k$ 矩阵, $B$ 是 $k\times n$ 矩阵, 则$$(AB{)}^T=B^T{A}^T$$类似地, 对于多个矩阵的乘积, $(A_1{A}_2\cdots A_l{)}^T=A_l^T\cdots A_2^T{A}_1^T$.

证明: 留作练习.

$\square$

定理 3.3.11. $n$ 阶行列式具有转置不变性, 即 $\det A=\det A^T$.

证明: 当 $n=1$ 时显然成立, 我们对 $n$ 做归纳. 对 $\det A$ 作行展开, 对 $\det A^T$ 作列展开, 得到$$\det A=\sum _{j=1}^n(-1{)}^{1+j}{a}_{1j}{A}_{1j}\det A^T=\sum _{i=1}^n(-1{)}^{1+i}{a}_{1i}{A}_{i1}^T$$由归纳假设知 $A_{1i}=A_{i1}^T$, 因此 $\det A=\det A^T$.

$\square$

转置不变性说明在行列式中行和列的地位是平等的. 因此行列式关于行变换的性质对于列变换也成立. 例如我们有

•	关于任意一列具有线性性$$\left|\begin{array}{ccc}\cdots & a_1+b_1& \cdots \\ \cdots & a_2+b_2& \cdots \\ \cdots & ⋮& \cdots \\ \cdots & a_n+b_n& \cdots \end{array}\right|=\left|\begin{array}{ccc}\cdots & a_1& \cdots \\ \cdots & a_2& \cdots \\ \cdots & ⋮& \cdots \\ \cdots & a_n& \cdots \end{array}\right|+\left|\begin{array}{ccc}\cdots & b_1& \cdots \\ \cdots & b_2& \cdots \\ \cdots & ⋮& \cdots \\ \cdots & b_n& \cdots \end{array}\right|$$$$\left|\begin{array}{ccccc}{a}_{11}& \cdots & \lambda a_{1j}& \cdots & a_{1n}\\ a_{21}& \cdots & \lambda a_{2j}& \cdots & a_{2n}\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{n1}& \cdots & \lambda a_{nj}& \cdots & a_{nn}\end{array}\right|=\lambda \left|\begin{array}{ccccc}{a}_{11}& \cdots & a_{1j}& \cdots & a_{1n}\\ a_{21}& \cdots & a_{2j}& \cdots & a_{2n}\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{n1}& \cdots & a_{nj}& \cdots & a_{nn}\end{array}\right|$$
•	交换两列, 行列式差一个负号$$\left|\begin{array}{ccccccc}{a}_{11}& \cdots & a_{1i}& \cdots & a_{1j}& \cdots & a_{1n}\\ a_{21}& \cdots & a_{2i}& \cdots & a_{2j}& \cdots & a_{2n}\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{n1}& \cdots & a_{ni}& \cdots & a_{nj}& \cdots & a_{nn}\end{array}\right|=-\left|\begin{array}{ccccccc}{a}_{11}& \cdots & a_{1j}& \cdots & a_{1i}& \cdots & a_{1n}\\ a_{21}& \cdots & a_{2j}& \cdots & a_{2i}& \cdots & a_{2n}\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{n1}& \cdots & a_{nj}& \cdots & a_{ni}& \cdots & a_{nn}\end{array}\right|$$
•	把一列乘以一个数加到另一列, 行列式不变$$\left|\begin{array}{ccccccc}{a}_{11}& \cdots & a_{1i}+\lambda a_{1j}& \cdots & a_{1j}& \cdots & a_{1n}\\ a_{21}& \cdots & a_{2i}+\lambda a_{2j}& \cdots & a_{2j}& \cdots & a_{2n}\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{n1}& \cdots & a_{ni}+\lambda a_{nj}& \cdots & a_{nj}& \cdots & a_{nn}\end{array}\right|=-\left|\begin{array}{ccccccc}{a}_{11}& \cdots & a_{1i}& \cdots & a_{1j}& \cdots & a_{1n}\\ a_{21}& \cdots & a_{2i}& \cdots & a_{2j}& \cdots & a_{2n}\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{n1}& \cdots & a_{ni}& \cdots & a_{nj}& \cdots & a_{nn}\end{array}\right|$$

一些例子

例子 3.3.12 (Vandermonde 行列式). 如下 $n$ 阶行列式$${\Delta }_n(x_1,\cdots ,x_n):=\left|\begin{array}{cccc}1& 1& \cdots & 1\\ x_1& x_2& \cdots & x_n\\ x_1^2& x_2^2& \cdots & x_n^2\\ ⋮& ⋮& ⋮& ⋮\\ x_1^{n-1}& x_2^{n-1}& \cdots & x_n^{n-1}\end{array}\right|$$称为 Vandermonde 行列式, 在数学和物理中有很多应用.

我们用展开定理来计算 ${\Delta }_n(x_1,\cdots ,x_n)$. 把第 $(n-1)$ 行乘以 $-x_1$ 加到第 $n$ 行, 然后把第 $(n-2)$ 行乘以 $-x_1$ 加到第 $n-1$ 行, $\cdots$, 最后把第 $1$ 行乘以 $-x_1$ 加到第 $2$ 行, 我们得到$$\begin{array}{r}\left|\begin{array}{cccc}1& 1& \cdots & 1\\ x_1& x_2& \cdots & x_n\\ x_1^2& x_2^2& \cdots & x_n^2\\ ⋮& ⋮& ⋮& ⋮\\ x_1^{n-1}& x_2^{n-1}& \cdots & x_n^{n-1}\end{array}\right|=\left|\begin{array}{cccc}1& 1& \cdots & 1\\ 0& x_2-x_1& \cdots & x_n-x_1\\ 0& (x_2-x_1)x_2& \cdots & (x_n-x_1)x_n\\ ⋮& ⋮& ⋮& ⋮\\ 0& (x_2-x_1)x_2^{n-2}& \cdots & (x_n-x_1)x_n^{n-2}\end{array}\right|\end{array}$$按照第 $1$ 列展开, 我们得到$$\begin{array}{rl}=& \left|\begin{array}{ccc}{x}_2-x_1& \cdots & x_n-x_1\\ (x_2-x_1)x_2& \cdots & (x_n-x_1)x_n\\ ⋮& ⋮& ⋮\\ (x_2-x_1)x_2^{n-2}& \cdots & (x_n-x_1)x_n^{n-2}\end{array}\right|\\ =& (x_2-x_1)(x_3-x_1)\cdots (x_n-x_1)\left|\begin{array}{cccc}1& 1& \cdots & 1\\ x_2& x_3& \cdots & x_n\\ ⋮& ⋮& ⋮& ⋮\\ x_2^{n-2}& x_3^{n-2}& \cdots & x_n^{n-2}\end{array}\right|\end{array}$$这里第二个等式是因为提取了每列的共同系数. 因此我们得到递推关系$${\Delta }_n(x_1,\cdots ,x_n)={\Delta }_{n-1}(x_2,\cdots ,x_n)\prod _{i=2}^n(x_i-x_1).$$由此得到$$\begin{array}{rl}& {\Delta }_n(x_1,\cdots ,x_n)={\Delta }_{n-1}(x_2,\cdots ,x_n)\prod _{i=2}^n(x_i-x_1)\\ =& {\Delta }_{n-2}(x_3,\cdots ,x_n)\left(\prod _{i=3}^n(x_i-x_2)\right)\left(\prod _{i=2}^n(x_i-x_1)\right)\\ =& \cdots =\prod _{i>j}(x_i-x_j)\end{array}$$

例子 3.3.13. 求如下 $n$ 阶三对角矩阵的行列式$${\Delta }_n={\left|\begin{array}{cccccc}-2& 1& 0& 0& \cdots & 0\\ 1& -2& 1& 0& \cdots & 0\\ 0& 1& -2& 1& \cdots & 0\\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0& 0& 0& \cdots & 1& -2\end{array}\right|}_n$$这里右下角标 ${|}_n$ 表示是 $n$ 阶方阵. 这个矩阵在 Poisson 方程边值问题的数值解法中扮演了十分重要的角色. 把行列式通过第一行展开, 我们得到$$\begin{array}{rl}& {\left|\begin{array}{cccccc}-2& 1& 0& 0& \cdots & 0\\ 1& -2& 1& 0& \cdots & 0\\ 0& 1& -2& 1& \cdots & 0\\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0& 0& 0& \cdots & 1& -2\end{array}\right|}_n\\ =& （-2）{\left|\begin{array}{ccccc}-2& 1& 0& \cdots & 0\\ 1& -2& 1& \cdots & 0\\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 0& 0& \cdots & 1& -2\end{array}\right|}_{n-1}-{\left|\begin{array}{ccccc}1& 1& 0& \cdots & 0\\ 0& -2& 1& \cdots & 0\\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 0& 0& \cdots & 1& -2\end{array}\right|}_{n-1}\end{array}$$

再把上述第二个矩阵按照第一列展开$$\begin{array}{rl}=& （-2）{\left|\begin{array}{ccccc}-2& 1& 0& \cdots & 0\\ 1& -2& 1& \cdots & 0\\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 0& 0& \cdots & 1& -2\end{array}\right|}_{n-1}-{\left|\begin{array}{cccc}-2& 1& \cdots & 0\\ \cdots & \cdots & \cdots & \cdots \\ 0& \cdots & 1& -2\end{array}\right|}_{n-2}\end{array}$$因此我们得到递推关系$${\Delta }_n=-2{\Delta }_{n-1}-{\Delta }_{n-2}$$初始条件为 ${\Delta }_1=-2,{\Delta }_2=3$.

我们把递推关系写成 $({\Delta }_n+{\Delta }_{n-1})=-({\Delta }_{n-1}+{\Delta }_{n-2})$, 因此$$\begin{array}{rl}& ({\Delta }_n+{\Delta }_{n-1})=-({\Delta }_{n-1}+{\Delta }_{n-2})\\ =& \cdots =(-1{)}^{n-2}({\Delta }_2+{\Delta }_1)=(-1{)}^n\end{array}$$代入得$$\begin{array}{rl}{\Delta }_n=& (-1{)}^n-{\Delta }_{n-1}\\ =& (-1{)}^n-(-1{)}^{n-1}+{\Delta }_{n-2}=\cdots \\ =& (-1{)}^n-(-1{)}^{n-1}+\cdots +(-1{)}^2+(-1{)}^{n-1}{\Delta }_1\\ =& (n+1)(-1{)}^n\end{array}$$