高専の数学解説

第17章 17.22

(1)

∣A−λE∣=∣2−λ0102−λ1111−λ∣=(2−λ)2(1−λ)−2(2−λ)=(2−λ)(2−3λ+λ2−2)|A - \lambda E| = \begin{vmatrix} 2-\lambda & 0 & 1 \\ 0 & 2-\lambda & 1 \\ 1 & 1 & 1-\lambda \end{vmatrix} = (2-\lambda)^2(1-\lambda) - 2(2-\lambda) = (2-\lambda)(2-3\lambda+\lambda^2-2)

=(2−λ)(λ2−3λ)=λ(2−λ)(λ−3)=0λ=0,2,3= (2-\lambda)(\lambda^2-3\lambda) = \lambda(2-\lambda)(\lambda-3) = 0 \quad \lambda = 0,2,3

(2)

λ=0\lambda = 0のとき

(201021111)→(111021201)→(1110210−2−1)→(111021000)→(222021000)→(201021000)\begin{pmatrix} 2 & 0 & 1 \\ 0 & 2 & 1 \\ 1 & 1 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 2 & 0 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & -2 & -1 \end{pmatrix} \to \begin{pmatrix} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 2 & 2 & 2 \\ 0 & 2 & 1 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 2 & 0 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & 0 \end{pmatrix}

p1=c1(−12)→単位固有ベクトル16(1−2)\mathbf{p}_1 = c_1 \begin{pmatrix} -1 \\ 2 \end{pmatrix} \to \text{単位固有ベクトル} \frac{1}{\sqrt{6}} \begin{pmatrix} 1 \\ -2 \end{pmatrix}

λ=2\lambda = 2のとき

(00100111−1)→(11−1001000)→(110001000)\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & -1 \end{pmatrix} \to \begin{pmatrix} 1 & 1 & -1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}

p2=c2(−10)→単位固有ベクトル12(−10)\mathbf{p}_2 = c_2 \begin{pmatrix} -1 \\ 0 \end{pmatrix} \to \text{単位固有ベクトル} \frac{1}{\sqrt{2}} \begin{pmatrix} -1 \\ 0 \end{pmatrix}

λ=3\lambda = 3のとき

(−1010−1111−2)→(10−10−1111−2)→(10−10−1101−1)→(10−101−1000)\begin{pmatrix} -1 & 0 & 1 \\ 0 & -1 & 1 \\ 1 & 1 & -2 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & -1 \\ 0 & -1 & 1 \\ 1 & 1 & -2 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & -1 \\ 0 & -1 & 1 \\ 0 & 1 & -1 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}

p3=c3(11)→単位固有ベクトル13(11)\mathbf{p}_3 = c_3 \begin{pmatrix} 1 \\ 1 \end{pmatrix} \to \text{単位固有ベクトル} \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ 1 \end{pmatrix}

これから

P=(16121316−1213−26013)P−1AP=(000020003)P = \begin{pmatrix} \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\ -\frac{2}{\sqrt{6}} & 0 & \frac{1}{\sqrt{3}} \end{pmatrix} \quad P^{-1}AP = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}

(3)

An=P(00002n0003n)P−1=(16121316−1213−26013)(00002n0003n)(1616−2612−120131313)A^n = P \begin{pmatrix} 0 & 0 & 0 \\ 0 & 2^n & 0 \\ 0 & 0 & 3^n \end{pmatrix} P^{-1} = \begin{pmatrix} \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\ -\frac{2}{\sqrt{6}} & 0 & \frac{1}{\sqrt{3}} \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \\ 0 & 2^n & 0 \\ 0 & 0 & 3^n \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \end{pmatrix}

=(02n23n30−2n23n3003n3)(1616−2612−120131313)=(2n−1+3n−11−2n−1+3n−13n−1−2n−1+3n−12n−1+3n−13n−13n−113n−113n−11)= \begin{pmatrix} 0 & \frac{2^n}{\sqrt{2}} & \frac{3^n}{\sqrt{3}} \\ 0 & -\frac{2^n}{\sqrt{2}} & \frac{3^n}{\sqrt{3}} \\ 0 & 0 & \frac{3^n}{\sqrt{3}} \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \end{pmatrix} = \begin{pmatrix} \frac{2^{n-1}+3^{n-1}}{1} & -2^{n-1}+3^{n-1} & 3^{n-1} \\ -2^{n-1}+3^{n-1} & 2^{n-1}+3^{n-1} & 3^{n-1} \\ \frac{3^{n-1}}{1} & \frac{3^{n-1}}{1} & \frac{3^{n-1}}{1} \end{pmatrix}

解説: r31bn1z