高専の数学解説

第10章 10.21

(1)

A=(5813)A = \begin{pmatrix} 5 & 8 \\ 1 & 3 \end{pmatrix}

∣A−λE∣=∣5−λ813−λ∣=(5−λ)(3−λ)−8=λ2−8λ+7=0=(λ−7)(λ−1)=0|A - \lambda E| = \begin{vmatrix} 5-\lambda & 8 \\ 1 & 3-\lambda \end{vmatrix} = (5-\lambda)(3-\lambda) - 8 = \lambda^2 - 8\lambda + 7 = 0 = (\lambda - 7)(\lambda - 1) = 0

∴λ=7,1\therefore \lambda = 7, 1

λ=7\lambda = 7 のとき

(5−7813−7)(xy)=(00)\begin{pmatrix} 5-7 & 8 \\ 1 & 3-7 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}

{−2x+8y=0x−4y=0x=4yよってP1=(41)\begin{cases} -2x + 8y = 0 \\ x - 4y = 0 \end{cases} \quad x = 4y \quad \text{よって} \quad P_1 = \begin{pmatrix} 4 \\ 1 \end{pmatrix}

λ=1\lambda = 1 のとき

(5−1813−1)(xy)=(00)\begin{pmatrix} 5-1 & 8 \\ 1 & 3-1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}

{4x+8y=0x+2y=0x=−2yよってP2=(2−1)\begin{cases} 4x + 8y = 0 \\ x + 2y = 0 \end{cases} \quad x = -2y \quad \text{よって} \quad P_2 = \begin{pmatrix} 2 \\ -1 \end{pmatrix}

これから 固有値 λ1=7,λ2=1\lambda_1 = 7, \lambda_2 = 1 固有ベクトル P1=(41),P2=(2−1)P_1 = \begin{pmatrix} 4 \\ 1 \end{pmatrix}, P_2 = \begin{pmatrix} 2 \\ -1 \end{pmatrix}

(1)

から P1=(41),P2=(2−1)P_1 = \begin{pmatrix} 4 \\ 1 \end{pmatrix}, P_2 = \begin{pmatrix} 2 \\ -1 \end{pmatrix}

P=(24−11)P−1=16(1−412)P = \begin{pmatrix} 2 & 4 \\ -1 & 1 \end{pmatrix} \quad P^{-1} = \frac{1}{6} \begin{pmatrix} 1 & -4 \\ 1 & 2 \end{pmatrix}

P−1AP=16(1−412)(5813)(24−11)=(1007)P^{-1}AP = \frac{1}{6} \begin{pmatrix} 1 & -4 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} 5 & 8 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 2 & 4 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 7 \end{pmatrix}

(3)

16(1−412)(x(0)y(0))=(X(0)Y(0))\frac{1}{6} \begin{pmatrix} 1 & -4 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} x_{(0)} \\ y_{(0)} \end{pmatrix} = \begin{pmatrix} X_{(0)} \\ Y_{(0)} \end{pmatrix}

(X(0)Y(0))=16(1−412)(11)=12(−11)\begin{pmatrix} X_{(0)} \\ Y_{(0)} \end{pmatrix} = \frac{1}{6} \begin{pmatrix} 1 & -4 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} -1 \\ 1 \end{pmatrix}

P−1(x(0)y(0))=(X(0)Y(0))から(x(0)y(0))=P(X(0)Y(0))P^{-1} \begin{pmatrix} x_{(0)} \\ y_{(0)} \end{pmatrix} = \begin{pmatrix} X_{(0)} \\ Y_{(0)} \end{pmatrix} \quad \text{から} \quad \begin{pmatrix} x_{(0)} \\ y_{(0)} \end{pmatrix} = P \begin{pmatrix} X_{(0)} \\ Y_{(0)} \end{pmatrix}

(x(0)′y(0)′)=A(x(0)y(0))=AP(X(0)Y(0))\begin{pmatrix} x'_{(0)} \\ y'_{(0)} \end{pmatrix} = A \begin{pmatrix} x_{(0)} \\ y_{(0)} \end{pmatrix} = AP \begin{pmatrix} X_{(0)} \\ Y_{(0)} \end{pmatrix}

P−1(x(0)′y(0)′)=P−1AP(X(0)Y(0))P^{-1} \begin{pmatrix} x'_{(0)} \\ y'_{(0)} \end{pmatrix} = P^{-1}AP \begin{pmatrix} X_{(0)} \\ Y_{(0)} \end{pmatrix}

(X′Y′)=(1007)(XY)\begin{pmatrix} X' \\ Y' \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 7 \end{pmatrix} \begin{pmatrix} X \\ Y \end{pmatrix}

(3)

から X′=X①X' = X \quad \text{①} Y′=7Y②Y' = 7Y \quad \text{②}

X(0)=−12,Y(0)=12X_{(0)} = -\frac{1}{2}, \quad Y_{(0)} = \frac{1}{2}

①から dXdt=X\frac{dX}{dt} = X ∫1XdX=∫dt\int \frac{1}{X} dX = \int dt log⁡X=t+C1\log X = t + C_1 eC1=C2 とおくとe^{C_1} = C_2 \text{ とおくと}

X=AetX = Ae^t

X(0)=−12 から A=−12 これからX_{(0)} = -\frac{1}{2} \text{ から } A = -\frac{1}{2} \text{ これから}

X=−12etX = -\frac{1}{2}e^t

②から同様に考えて

Y=C3etY = C_3 e^t

Y(0)=12 から B=12 これからY_{(0)} = \frac{1}{2} \text{ から } B = \frac{1}{2} \text{ これから}

Y=12e7tY = \frac{1}{2}e^{7t}

(x(t)y(t))=p(X(t)Y(t))=(24−11)(−12et12e7t)=12(−2et+4e7tet+e7t)\begin{pmatrix} x_{(t)} \\ y_{(t)} \end{pmatrix} = p \begin{pmatrix} X_{(t)} \\ Y_{(t)} \end{pmatrix} = \begin{pmatrix} 2 & 4 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} -\frac{1}{2}e^t \\ \frac{1}{2}e^{7t} \end{pmatrix} = \frac{1}{2} \begin{pmatrix} -2e^t + 4e^{7t} \\ e^t + e^{7t} \end{pmatrix}

解説: r31bn1z