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# 如何用递推法求 $$n$$ 阶行列式

$D_{2 n} = \left|\begin{array}{cccccccc} a_n & & & & & & & b_n\\ 0 & a_{n – 1} & & & & & b_{n – 1} & 0\\ & & \ddots & & &\cdot^{\cdot^{\cdot}} & & \\ & & & a_1 & b_1 & & & \\ & & & c_1 & d_1 & & & \\ & &\cdot^{\cdot^{\cdot}} & & & \ddots & & \\ 0 & c_{n – 1} & & & & & d_{n – 1} & 0\\ c_n & & & & & & & d_n \end{array}\right|$

\begin{align*} D_{2 n} &= a_n \left|\begin{array}{ccccccc}
a_{n – 1} & & & & & b_{n – 1} & 0\\
& \ddots & & & \cdot^{\cdot^{\cdot}} & & \\
& & a_1 & b_1 & & & \\
& & c_1 & d_1 & & & \\
& \cdot^{\cdot^{\cdot}} & & & \ddots & & \\
c_{n – 1} & & & & & d_{n – 1} & \\
0 & & & & & & d_n
\end{array}\right|\\
& + ( – 1)^{1 + 2 n} b_n \left|\begin{array}{ccccccc}
0 & a_{n – 1} & & & & & b_{n – 1}\\
& & \ddots & & & \cdot^{\cdot^{\cdot}} & \\
& & & a_1 & b_1 & & \\
& & & c_1 & d_1 & & \\
& &\cdot^{\cdot^{\cdot}} & & & \ddots & \\
0 & c_{n – 1} & & & & & d_{n – 1}\\
c_n & & & & & & 0
\end{array}\right|,\end{align*}

\begin{align*} D_{2 n} &= a_n d_n \left|\begin{array}{cccccc}
a_{n – 1} & & & & & b_{n – 1}\\
& \ddots & & & \cdot^{\cdot^{\cdot}} & \\
& & a_1 & b_1 & & \\
& & c_1 & d_1 & & \\
&\cdot^{\cdot^{\cdot}} & & & \ddots & \\
c_{n – 1} & & & & & d_{n – 1}
\end{array}\right|\\
& + ( – 1)^{1 + 2 n} ( – 1)^{1 + 2 n – 1} b_n c_n
\left|\begin{array}{cccccc}
a_{n – 1} & & & & & b_{n – 1}\\
& \ddots & & & \cdot^{\cdot^{\cdot}} & \\
& & a_1 & b_1 & & \\
& & c_1 & d_1 & & \\
& \cdot^{\cdot^{\cdot}} & & & \ddots & \\
c_{n – 1} & & & & & d_{n – 1}
\end{array}\right|, \end{align*}

$D_{2 ( n – 1)} = \left|\begin{array}{cccccc} a_{n – 1} & & & & & b_{n – 1}\\ & \ddots & & & \cdot^{\cdot^{\cdot}} & \\ & & a_1 & b_1 & & \\ & & c_1 & d_1 & & \\ & \cdot^{\cdot^{\cdot}} & & & \ddots & \\ c_{n – 1} & & & & & d_{n – 1} \end{array}\right|,$

$D_{2 n} = ( a_n d_n – b_n c_n) D_{2 ( n – 1)} .$

$D_{2 ( n – 1)} = ( a_{n – 1} d_{n – 1} – b_{n – 1} c_{n – 1}) D_{2 ( n – 2)}, D_{2 ( n – 2)} = ( a_{n – 2} d_{n – 2} – b_{n – 2} c_{n – 2}) D_{2 ( n – 3)}$

$D_{2 n} = ( a_n d_n – b_n c_n) ( a_{n – 1} d_{n – 1} – b_{n – 1} c_{n – 1}) \cdots ( a_2 d_2 – b_2 d_2) D_2,$

$D_{2 n} = ( a_n d_n – b_n c_n) ( a_{n – 1} d_{n – 1} – b_{n – 1} c_{n – 1}) \cdots ( a_1 d_1 – b_1 d_1) .$