例题 5.5

例题 5.5.

证明: $n + k$ 阶行列式 $D = a_{11} a_{21} ⋮ a_{k 1} b_{11} b_{21} ⋮ b_{n 1} a_{12} a_{22} ⋮ a_{k 2} b_{12} b_{22} ⋮ b_{n 2} \dots \dots \dots \dots \dots \dots a_{1 k} a_{2 k} ⋮ a_{kk} b_{1 k} b_{2 k} ⋮ b_{nk} 00 ⋮ 0 c_{11} c_{21} ⋮ c_{n 1} 00 ⋮ 0 c_{12} c_{22} ⋮ c_{n 2} \dots \dots \dots \dots \dots \dots 00 ⋮ 0 c_{1 n} c_{2 n} ⋮ c_{nn} = D_{1} D_{2},$ 其中 $D_{1} = a_{11} a_{21} ⋮ a_{k 1} a_{12} a_{22} ⋮ a_{k 2} \dots \dots \dots a_{1 k} a_{2 k} ⋮ a_{kk}, D_{2} = c_{11} c_{21} ⋮ c_{n 1} c_{12} c_{22} ⋮ c_{n 2} \dots \dots \dots c_{1 n} c_{2 n} ⋮ c_{nn} .$

\begin{proof} 由于 $D$ 来自前 $k$ 行的 $k$ 阶子式只有一个, 即左上角的 $k$ 阶子式 $D_{1}$ 可能不为零. 此时, 由拉普拉斯展开定理知 $D = (- 1)^{1 + 2 + \dots + k + 1 + 2 + \dots + k} D_{1} D_{2} = D_{1} D_{2} .$ \end{proof}

Youliang Zhong

例题 5.5

Graph View

Backlinks