高専の数学解説

第17章 17.18

(1)

∣P∣=∣101−1121−1−1∣=−1+1−1+2=1|P|=\begin{vmatrix}1 & 0 & 1\\-1 & 1 & 2\\1 & -1 & -1\end{vmatrix}=-1+1-1+2=1

∣p11∣=∣12−1−1∣=−1+2=1∣p12∣=∣−121−1∣=1−2=−1∣p13∣=∣−111−1∣=1−1=0|p_{11}|=\begin{vmatrix}1 & 2\\-1 & -1\end{vmatrix}=-1+2=1 \quad |p_{12}|=\begin{vmatrix}-1 & 2\\1 & -1\end{vmatrix}=1-2=-1 \quad |p_{13}|=\begin{vmatrix}-1 & 1\\1 & -1\end{vmatrix}=1-1=0

∣p21∣=∣01−1−1∣=1∣p22∣=∣111−1∣=−1−1=−2∣p23∣=∣101−1∣=−1|p_{21}|=\begin{vmatrix}0 & 1\\-1 & -1\end{vmatrix}=1 \quad |p_{22}|=\begin{vmatrix}1 & 1\\1 & -1\end{vmatrix}=-1-1=-2 \quad |p_{23}|=\begin{vmatrix}1 & 0\\1 & -1\end{vmatrix}=-1

∣p31∣=∣0112∣=−1∣p32∣=∣11−12∣=2+1=3∣p33∣=∣10−11∣=1|p_{31}|=\begin{vmatrix}0 & 1\\1 & 2\end{vmatrix}=-1 \quad |p_{32}|=\begin{vmatrix}1 & 1\\-1 & 2\end{vmatrix}=2+1=3 \quad |p_{33}|=\begin{vmatrix}1 & 0\\-1 & 1\end{vmatrix}=1

これから

P−1=(1−1−11−2−3011)P^{-1}=\begin{pmatrix}1 & -1 & -1\\1 & -2 & -3\\0 & 1 & 1\end{pmatrix}

(2)

∣A−λE∣=∣−1−λ3321−λ0−212−λ∣=−(1+λ)(1−λ)(2−λ)+6+6(1−λ)−6(2−λ)|A-\lambda E|=\begin{vmatrix}-1-\lambda & 3 & 3\\2 & 1-\lambda & 0\\-2 & 1 & 2-\lambda\end{vmatrix}=-(1+\lambda)(1-\lambda)(2-\lambda)+6+6(1-\lambda)-6(2-\lambda)

=(1−λ)(λ2−λ+4)−6(1−λ)=(1−λ)(λ2−λ+4−6)=(1−λ)(λ+1)(λ−2)λ=1,2,−1=(1-\lambda)(\lambda^2-\lambda+4)-6(1-\lambda)=(1-\lambda)(\lambda^2-\lambda+4-6)=(1-\lambda)(\lambda+1)(\lambda-2) \quad \lambda=1,2,-1

λ=1\lambda=1のとき

(−233200−211)→(200−233−211)→(200022011)→(100011000)p1=C1(01−1)\begin{pmatrix}-2 & 3 & 3\\2 & 0 & 0\\-2 & 1 & 1\end{pmatrix}\rightarrow\begin{pmatrix}2 & 0 & 0\\-2 & 3 & 3\\-2 & 1 & 1\end{pmatrix}\rightarrow\begin{pmatrix}2 & 0 & 0\\0 & 2 & 2\\0 & 1 & 1\end{pmatrix}\rightarrow\begin{pmatrix}1 & 0 & 0\\0 & 1 & 1\\0 & 0 & 0\end{pmatrix} \quad p_1=C_1\begin{pmatrix}0\\1\\-1\end{pmatrix}

λ=2\lambda=2のとき

(−3332−10−210)→(1−1−12−10000)→(1−1−1012000)→(101012000),p2=C2(12−1)\begin{pmatrix} -3 & 3 & 3 \\ 2 & -1 & 0 \\ -2 & 1 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & -1 & -1 \\ 2 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & -1 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{pmatrix}, \quad p_2 = C_2 \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}

λ=−1\lambda = -1 のとき

(033220−213)→(110033−213)→(110011033)→(110011000)→(10−1011000),p3=C3(1−11)\begin{pmatrix} 0 & 3 & 3 \\ 2 & 2 & 0 \\ -2 & 1 & 3 \end{pmatrix} \to \begin{pmatrix} 1 & 1 & 0 \\ 0 & 3 & 3 \\ -2 & 1 & 3 \end{pmatrix} \to \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 3 & 3 \end{pmatrix} \to \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}, \quad p_3 = C_3 \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}

これから

(−100010002)\begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}

(3)

A5=(101−1121−1−1)((−1)50001500025)(1−1−11−2−3011)=(101−1121−1−1)(−1000100032)(1−1−11−2−3011)A^5 = \begin{pmatrix} 1 & 0 & 1 \\ -1 & 1 & 2 \\ 1 & -1 & -1 \end{pmatrix} \begin{pmatrix} (-1)^5 & 0 & 0 \\ 0 & 1^5 & 0 \\ 0 & 0 & 2^5 \end{pmatrix} \begin{pmatrix} 1 & -1 & -1 \\ 1 & -2 & -3 \\ 0 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 1 \\ -1 & 1 & 2 \\ 1 & -1 & -1 \end{pmatrix} \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 32 \end{pmatrix} \begin{pmatrix} 1 & -1 & -1 \\ 1 & -2 & -3 \\ 0 & 1 & 1 \end{pmatrix}

=(−10321164−1−1−32)(1−1−11−2−3011)=(−1333326160−2−29−28)= \begin{pmatrix} -1 & 0 & 32 \\ 1 & 1 & 64 \\ -1 & -1 & -32 \end{pmatrix} \begin{pmatrix} 1 & -1 & -1 \\ 1 & -2 & -3 \\ 0 & 1 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 33 & 33 \\ 2 & 61 & 60 \\ -2 & -29 & -28 \end{pmatrix}

解説: r31bn1z