3.2 解答

3.2

1.

(a) $\det(A - \lambda I) = \left(\begin{array}{cc} 6-\lambda&8\\ -1&2-\lambda \end{array}\right) = \lambda^2 - 8\lambda + 20 = 0$. よって固有値は $\lambda = 4 \pm 2i$である．次に固有値 $\lambda = 4 + 2i$に対する固有ベクトル${\bf C}$は

$$(A - (4 + 2i)I){\bf C} = \left(\begin{array}{cc} 2 - 2i & 8\\ -1... ...\ c_{2} \end{array}\right) = \left(\begin{array}{c} 0\\ 0 \end{array}\right)$$

をみたす．ここで行列 $\left(\begin{array}{cc} 2 - 2i & 8\\ -1 & -2 - 2i \end{array}\right) \Longrightarrow \left(\begin{array}{cc} 1&2 + 2i\\ 0&0 \end{array}\right)$は被約階段行列に変形されるので， $c_{2} = \alpha$とおくと， $c_{1} = -2(1 + i)\alpha$となる．したがって，固有ベクトルは $\left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) = \alpha \left(\begin{array}{c} -2(1 + i) \\ 1 \end{array}\right)$である．これより ${\bf C}e^{\lambda t}$の実部と虚部を求めると

$${\bf C}e^{\lambda t} = \left(\begin{array}{c} -2(1 + i) \\ 1 \end{array}\right) e^{(4 +... ...\begin{array}{c} -2(1 + i) \\ 1 \end{array}\right)e^{4t}(\cos{2t} + i\sin{2t})$$

$$= \left(\begin{array}{c} -2(1+i)e^{4t}(\cos{2t} + i\sin{2t}) \\ e^{4t}(\cos{2t} + i\sin{2t}) \end{array}\right)$$

$$= \left(\begin{array}{c} -2e^{4t}\cos{2t} + 2e^{4t}\sin{2t}) \\ e^... ...y}{c} -2e^{4t}\cos{2t} - 2e^{4t}\sin{2t}) \\ e^{4t}\sin{2t} \end{array}\right)$$

これより一般解は次のように表わすことができる．

$${\bf X} = c_{1}\left(\begin{array}{c} -2\cos{2t} + 2\sin{2t}\\ \... ...c} -2\cos{2t} - 2\sin{2t}\\ \sin{2t} \end{array}\right)e^{4t} \ \ \framebox{終}$$

(b) $\det(A - \lambda I) = \left(\begin{array}{cc} 2-\lambda&-1\\ 4&6-\lambda \end{array}\right) = \lambda^2 - 8\lambda + 16 = (\lambda - 4)^{2}$ より固有値は $\lambda = 4$である．次に固有値 $\lambda = 4$に対する固有ベクトル${\bf C}$は

$$(A - 4I){\bf C} = \left(\begin{array}{cc} -2 & -1\\ 4 & 2 \end{a... ...\ c_{2} \end{array}\right) = \left(\begin{array}{c} 0\\ 0 \end{array}\right)$$

をみたす．

$$\left(\begin{array}{cc} -2&-1\\ 4&2 \end{array}\right) \Longrightarrow \left(\begin{array}{cc} 1&\frac{1}{2}\\ 0&0 \end{array}\right)$$

より $c_{2} = -2\alpha$とおくと， $c_{1} = \alpha$となる．したがって，固有ベクトルは $\left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) = \alpha \left(\begin{array}{c} 1 \\ -2 \end{array}\right)$である．次に，

$$(A - 4I)^2 {\bf C} = {\bf0}, \ (A - 4I){\bf C} \neq {\bf0}$$

を満たす${\bf C}$を見つけると

$$(A - 4I)^2 = \left(\begin{array}{cc} -2&-1\\ 4&2 \end{array}\right)^2 = \left(\begin{array}{cc} 0&0\\ 0&0 \end{array}\right)$$

より， ${\bf C} = \left(\begin{array}{c} \alpha\\ \beta \end{array}\right)$は $(A - 4I)^2 = {\bf0}$を満たす．ここでもう一つの条件を満たすように $\alpha,\beta$を選ぶと， ${\bf C} = \left(\begin{array}{c} 1\\ 1 \end{array}\right)$となり，2つめの解

$$e^{At}{\bf C} = e^{4t}e^{(A-4I)t}{\bf C} = e^{4t}[{\bf C} + t(A - 4I){\bf C} + \frac{t^2}{2!}(A - 4I)^2 {\bf C}]$$

$$= e^{4t}\{\left(\begin{array}{c} 1\\ 1 \end{array}\right) + \left(... ...array}\right)t\} = e^{4t}\left(\begin{array}{c} 1-3t\\ 1+6t \end{array}\right)$$

ここで $\left(\begin{array}{c} 1\\ -2 \end{array}\right), \left(\begin{array}{c} 1-3t\\ 1+6t \end{array}\right)$は1次独立なので，一般解は次のように表わすことができる．

$${\bf X} =e^{4t}[c_{1}\left(\begin{array}{c} 1\\ -2 \end{array... ...c_{2}\left(\begin{array}{c} 1-3t\\ 1+6t \end{array}\right)] \ \framebox{終}$$

(c)

${\bf X} = \left(\begin{array}{ccc} 5&2&2\\ 2&2&-4\\ 2&-4&2 \end{array}\right){\bf X}$より

$$\det(A - \lambda I) = \left(\begin{array}{ccc} 5-\lambda&2&2\\ 2&2-\lambda&-4\\ 2&-4&2-\lambda \end{array}\right)$$

$$= (5- \lambda)(\lambda^2 - 4\lambda -12) -2(-2\lambda + 12) + 2(2\lambda -12)$$

$$= (5-\lambda)(\lambda+2)(\lambda-6) + 4\lambda -24 + 4\lambda - 24$$

$$= (5-\lambda)(\lambda+2)(\lambda-6) + 8(\lambda - 6)$$

$$= (\lambda - 6)(-\lambda^2 + 3\lambda + 18)$$

$$= -(\lambda - 6)(\lambda^2 - 3\lambda - 18)$$

$$= - (\lambda - 6)(\lambda + 3)(\lambda - 6)$$

よって固有値は $\lambda = -3,6$である．
固有値 $\lambda = -3$に対する固有ベクトル${\bf C}$は $(A + 3I){\bf C} = {\bf0}$の0でない解より，Gaussの消去法を用いて求める．

$$A + 3I = \left(\begin{array}{ccc} 8&2&2\\ 2&5&-4\\ 2&-4&5 \end{array}\ri... ...in{array}{ccc} 1&\frac{1}{4}&\frac{1}{4}\\ 2&5&-4\\ 2&-4&5 \end{array}\right)$$

$$\longrightarrow \left(\begin{array}{ccc} 1&\frac{1}{4}&\frac{1}{4}\\ 0&\frac{9}{... ...gin{array}{ccc} 1&\frac{1}{4}&\frac{1}{4}\\ 0&1&-1\\ 0&0&0 \end{array}\right)$$

$$\longrightarrow \left(\begin{array}{ccc} 1&0&\frac{1}{2}\\ 0&1&-1\\ 0&0&0 \end{array}\right)$$

より $c_{3} = 2\alpha$とおくと，

$${\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3} \end{arr... ...d{array}\right) = \alpha\left(\begin{array}{c} -1\\ 2\\ 2 \end{array}\right)$$

これより解 ${\bf X}_{1} = \left(\begin{array}{c} -1\\ 2\\ 2 \end{array}\right)e^{-3t}$を得る．

次に，固有値 $\lambda = 6$に対する固有ベクトルを求める．

$$A - 6I = \left(\begin{array}{ccc} -1&2&2\\ 2&-4&-4\\ 2&-4&-4 \e... ...ghtarrow \left(\begin{array}{ccc} 1&-2&-2\\ 0&0&0\\ 0&0&0 \end{array}\right)$$

より自由度が2．よって $c_{2} = \beta, c_{3} = \gamma$とおくと，

$${\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3} \end{arr... ...{array}\right) + \gamma \left(\begin{array}{c} 2\\ 0\\ 1 \end{array}\right).$$

これより解

$${\bf X}_{2} = \left(\begin{array}{c} 2\\ 1\\ 0 \end{array}\righ... ...6t}, {\bf X}_{3} = \left(\begin{array}{c} 2\\ 0\\ 1 \end{array}\right)e^{6t}$$

を得る．ここで ${\bf X}_{1},{\bf X}_{2},{\bf X}_{3}$は互いに1次独立なので，一般解は

$${\bf X} = c_{1}\left(\begin{array}{c} -1\\ 2\\ 2 \end{array}\ri... ...ght)e^{6t} + c_{3}\left(\begin{array}{c} 2\\ 0\\ 1 \end{array}\right)e^{6t}.$$

で与えられる．

(d)

$$\det(A - \lambda I) = \left\vert\begin{array}{ccc} 1-\lambda&0&1\\ 1&1-\lambda&0\\ -2&0&-1-\lambda \end{array}\right\vert$$

$$= (1-\lambda)(\lambda^2 - 1) + (-2\lambda + 2)$$

$$= (1-\lambda)(\lambda^2 -1) + 2(1-\lambda)$$

$$= (1-\lambda)(\lambda^2 + 1)$$

より固有値 $\lambda = 1, \pm i$を得る．

固有値 $\lambda = 1$に対する固有ベクトル${\bf C}$をGaussの消去法を用いて求める．

$$A - I = \left(\begin{array}{ccc} 0&0&1\\ 1&0&0\\ -2&0&-2 \end{array}\ri... ...rightarrow \left(\begin{array}{ccc} 1&0&0\\ 0&0&1\\ 0&0&-2 \end{array}\right)$$

$$\longrightarrow \left(\begin{array}{ccc} 1&0&0\\ 0&0&1\\ 0&0&0 \end{array}\right)$$

より $c_{2} = \alpha$とおくと，

$${\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3} \end{arr... ...nd{array}\right) = \alpha \left(\begin{array}{c} 0\\ 1\\ 0 \end{array}\right)$$

よって解 $X_{1} = \left(\begin{array}{c} 0\\ 1\\ 0 \end{array}\right)e^{t}$を得る．

次に $\lambda = i$に対する固有ベクトル${\bf C}$を求める．

$$A - i I = \left(\begin{array}{ccc} 1-i&0&1\\ 1&1-i&0\\ -2&0&-1-i \end{arr... ...row \left(\begin{array}{ccc} 1&1-i&0\\ 1-i&0&1\\ -2&0&-1-i \end{array}\right)$$

$$\longrightarrow \left(\begin{array}{ccc} 1&1-i&0\\ 0&2i&1\\ 0&2-2i&-1-i \end{ar... ...eft(\begin{array}{ccc} 1&1-i&0\\ 0&1&- \frac{i}{2}\\ 0&0&0 \end{array}\right)$$

$$\longrightarrow \left(\begin{array}{ccc} 1&0&\frac{1+i}{2}\\ 0&1&- \frac{i}{2}\\ 0&0&0 \end{array}\right)$$

より $c_{3} = 2\alpha$とおくと，

$${\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3} \end{arr... ...rray}\right) = \alpha \left(\begin{array}{c} -1-i\\ i\\ 2 \end{array}\right)$$

ここで解は ${\bf X}_{2} = \Re {\bf C}e^{it}, {\bf X}_{3} = \Im {\bf C}e^{it}$で与えられるので， ${\bf C}e^{it}$の実部と虚部を求める．

$${\bf C}e^{it} = \left(\begin{array}{c} -1-i\\ i\\ 2 \end{array}\right)e^{it} = \left(\begin{array}{c} -1-i\\ i\\ 2 \end{array}\right)(\cos{t} + i \sin{t})$$

$$= \underbrace{\left(\begin{array}{c} -\cos{t} + \sin{t} \\ -\sin{t... ...{c} \cos{t} + \sin{t} \\ -\cos{t}\\ -2\sin{t} \end{array}\right)}_{\mbox{虚部}}$$

よって一般解は

$${\bf X} = c_{1}\left(\begin{array}{c} 0\\ 1\\ 0 \end{array}\rig... ...begin{array}{c} \cos{t} + \sin{t} \\ -\cos{t}\\ -2\sin{t} \end{array}\right)$$

で与えられる．

(e)

$$\left\{\begin{array}{rl} 4x_{1}^{\prime} + x_{1} + 2x_{2}^{\prime... ... -2x_{1}^{\prime} + 2x_{1} - 2x_{2}^{\prime} - 2x_{2}& = 0 \end{array}\right.$$

$x_{1}^{\prime}$について解くと， $2x_{1}^{\prime} + 3x_{1} + 5x_{2} = 0$より

$$x_{1}^{\prime} = - \frac{3}{2}x_{1} - \frac{5}{2}x_{2}$$ (A.8)

また $x_{2}^{\prime}$について解くと， $-2x_{2}^{\prime} + 5x_{1} + 3x_{2} = 0$より

$$x_{2}^{\prime} = \frac{5}{2}x_{1} + \frac{3}{2}x_{2}$$ (A.9)

これより

$${\bf X}^{\prime} = \left(\begin{array}{cc} -\frac{3}{2}&-\frac{5}{2}\\ \frac{5}{2}&\frac{3}{2} \end{array}\right){\bf X}$$

$$\det(A - \lambda I) = \left\vert\begin{array}{cc} -\frac{3}{2} - \lambda& - \frac{5}{2}... ...3}{2} - \lambda \end{array}\right\vert = \lambda^2 - \frac{9}{4} + \frac{25}{4}$$

$$= \lambda^2 + 4 = (\lambda + 2i)(\lambda - 2i)$$

より固有値は $\lambda = \pm 2i$．

固有値 $\lambda = 2i$に対する固有ベクトルをGaussの消去法を用いて求める．

$$A - 2i I = \left(\begin{array}{cc} -\frac{3}{2} - 2i & - \frac{5}{2}\\ \fra... ...}{2}& \frac{3}{2} - 2i \\ -\frac{3}{2} - 2i & - \frac{5}{2} \end{array}\right)$$

$$\longrightarrow \left(\begin{array}{cc} 1&\frac{1}{5}(3-4i)\\ 0&0 \end{array}\right)$$

より $c_{2} = \alpha$とおくと，

$${\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right... ...right) = -\frac{\alpha}{5}\left(\begin{array}{c} 3-4i\\ -5 \end{array}\right)$$

これより解 ${\bf X}_{1} = \Re {\bf C}e^{2it}, {\bf X}_{2} = \Im {\bf C}e^{2it}$を求める．

$${\bf C}e^{2it} = \left(\begin{array}{c} 3-4i\\ -5 \end{array}\right)e^{2it} = \left(\begin{array}{c} 3-4i\\ -5 \end{array}\right)(\cos{2t} + i\sin{2t})$$

$$= \underbrace{\left(\begin{array}{c} 3\cos{2t} + 4\sin{2t}\\ -5\co... ...n{array}{c} 3\sin{2t} - 4\cos{2t}\\ -5\sin{2t} \end{array}\right)}_{\mbox{虚部}}$$

よって一般解は

$${\bf X} = c_{1}\left(\begin{array}{c} 3\cos{2t} + 4\sin{2t}\\ -5... ... \left(\begin{array}{c} 3\sin{2t} - 4\cos{2t}\\ -5\sin{2t} \end{array}\right)$$

で与えられる．

(f)

$x_{2}^{\prime}$について解くと， $3x_{2}^{\prime} - 3x_{1} - 3x_{2} = 0$より

$$x_{2}^{\prime} = x_{1} + x_{2}$$

また $x_{1}^{\prime} - 2x_{1} + 5x_{2}^{\prime} = 0$より

$$x_{1}^{\prime} = 2x_{1} - 5x_{2}^{\prime} = - 3x_{1} - 5x_{2}$$

これより ${\bf X}^{\prime} = \left(\begin{array}{cc} -3&-5\\ 1&1 \end{array}\right){\bf X}.$となる．これより固有値と固有ベクトルを求める．

$$\det(A - \lambda I) = \left\vert\begin{array}{cc} -3-\lambda&-5\\ 1&1-\lambda \end{array}\right\vert = \lambda^2 + 2\lambda + 2 = 0$$

よって固有値 $\lambda = -1 \pm \sqrt{1-2} = -1 \pm i$を得る．

固有値 $\lambda = -1+i$に対する固有ベクトル${\bf C}$をGaussの消去法を用いて求める．

$$A + (1-i)I = \left(\begin{array}{cc} -2-i&-5\\ 1&2-i \end{array}... ...right) \longrightarrow \left(\begin{array}{cc} 1&2-i\\ 0&0 \end{array}\right)$$

より $c_{2} = \alpha$とおくと，

$${\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right... ...end{array}\right) = \alpha\left(\begin{array}{c} -2+i\\ 1 \end{array}\right).$$

これより解 ${\bf X}_{1} = \Re {\bf C}e^{(-1+i)t}, {\bf X}_{2} + \Im {\bf C}e^{(-1+i)t}$を求める．

$${\bf C}e^{(-1+i)t} = \left(\begin{array}{c} -2+i\\ 1 \end{array}\right)e^{-t}e^{it} = \left(\begin{array}{c} -2+i\\ 1 \end{array}\right)e^{-t}(\cos{t} + i\sin{t})$$

$$= e^{-t}\underbrace{\left(\begin{array}{c} -2\cos{t}-\sin{t}\\ \co... ...ft(\begin{array}{c} \cos{t}-2\sin{t}\\ \sin{t} \end{array}\right)}_{\mbox{虚部}}$$

よって一般解は

$${\bf X} = e^{-t}[c_{1}\left(\begin{array}{c} -2\cos{t}-\sin{t}\\ ... ...+ c_{2} \left(\begin{array}{c} \cos{t}-2\sin{t}\\ \sin{t} \end{array}\right)]$$

で与えられる．