例题 5.1

例题 5.1.

计算 $n$ 阶行列式 $D = a_{11} 00 ⋮ 0 a_{n 1} 000 ⋮ a_{n - 1, 2} a_{n 2} \dots \dots \dots \dots \dots 00 a_{3, n - 2} ⋮ 00 0 a_{2, n - 1} a_{3, n - 1} ⋮ 00 a_{1 n} a_{2 n} 0 ⋮ 00 .$

\begin{proof} 将行列式按第一行展开, 得

D = (- 1)^{1 + 1} a_{11} 00 ⋮ a_{n - 1, 2} a_{n 2} \dots \dots \dots \dots 0 a_{3, n - 2} ⋮ 00 a_{2, n - 1} a_{3, n - 1} ⋮ 00 a_{2 n} 0 ⋮ 00 + (- 1)^{1 + n} a_{1 n} 00 ⋮ 0 a_{n 1} 00 ⋮ a_{n - 1, 2} a_{n 2} \dots \dots \dots \dots 0 a_{3, n - 2} ⋮ 00 a_{2, n - 1} a_{3, n - 1} ⋮ 00 = a_{11} (- 1)^{\frac{1}{2} (n - 1) (n - 2)} a_{2 n} a_{3, n - 1} \dots a_{n 2} + (- 1)^{1 + n} a_{1 n} (- 1)^{\frac{1}{2} (n - 1) (n - 2)} a_{2, n - 1} a_{3, n - 2} \dots a_{n 1} = (- 1)^{\frac{1}{2} (n - 1) (n - 2)} a_{11} a_{2 n} a_{3, n - 1} \dots a_{n 2} + (- 1)^{\frac{1}{2} n (n - 1)} a_{1 n} a_{2, n - 1} a_{3, n - 2} \dots a_{n 1} .

\end{proof}

Youliang Zhong

例题 5.1

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