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数学の主張

P108 第4章 章末問題11(2)．

問題11(2)

　次の等式を証明せよ．

∣xa2a3⋯an1a1xa3⋯an1a1a2x⋯an1⋮⋮⋮⋱⋮⋮a1a2a3⋯x1a1a2a3⋯an1∣=∏i=1n(x−ai)\begin{vmatrix} x & a_2 & a_3 & \cdots & a_n & 1 \\ a_1 & x & a_3 & \cdots & a_n & 1 \\ a_1 & a_2 & x & \cdots & a_n & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ a_1 & a_2 & a_3 & \cdots & x & 1 \\ a_1 & a_2 & a_3 & \cdots & a_n & 1 \\ \end{vmatrix} = \prod_{i=1}^{n}(x-a_i)

解答

Section titled “解答”
∣xa2a3⋯an1a1xa3⋯an1a1a2x⋯an1⋮⋮⋮⋱⋮⋮a1a2a3⋯x1a1a2a3⋯an1∣=∣x−a100⋯000x−a20⋯0000x−a3⋯00⋮⋮⋮⋱⋮⋮000⋯x−an0a1a2a3⋯an1∣=(−1)2n+2∣x−a100⋯00x−a20⋯000x−a3⋯0⋮⋮⋮⋱⋮000⋯x−an∣=∏i=1n(x−ai)\begin{align} \begin{vmatrix} x & a_2 & a_3 & \cdots & a_n & 1 \\ a_1 & x & a_3 & \cdots & a_n & 1 \\ a_1 & a_2 & x & \cdots & a_n & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ a_1 & a_2 & a_3 & \cdots & x & 1 \\ a_1 & a_2 & a_3 & \cdots & a_n & 1 \\ \end{vmatrix} &= \begin{vmatrix} x-a_1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & x-a_2 & 0 & \cdots & 0 & 0 \\ 0 & 0 & x-a_3 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & x-a_n & 0 \\ a_1 & a_2 & a_3 & \cdots & a_n & 1 \\ \end{vmatrix} \\ &= (-1)^{2n+2} \begin{vmatrix} x-a_1 & 0 & 0 & \cdots & 0 \\ 0 & x-a_2 & 0 & \cdots & 0 \\ 0 & 0 & x-a_3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & x-a_n \\ \end{vmatrix} \\ &= \prod_{i=1}^{n}(x-a_i) \end{align}
  • (1)は1行目にn+1行目x(-1)を加える．ii行目にn+1x(-1)を加える．2≤i≤n2 \le i \le n．
  • (2)はn+1n+1列目で余因子展開をした．
  • (3)は対角行列式を展開し，整理した．