高専の数学解説

第17章 17.9

(1)

∣A−λE∣=∣−1−λ8−27−λ∣=−(1+λ)(7−λ)+16=−(7+6λ−λ2)+16=λ2−6λ+9=(λ−3)2=0|A - \lambda E| = \begin{vmatrix} -1-\lambda & 8 \\ -2 & 7-\lambda \end{vmatrix} = -(1+\lambda)(7-\lambda) + 16 = -(7 + 6\lambda - \lambda^2) + 16 = \lambda^2 - 6\lambda + 9 = (\lambda-3)^2 = 0

λ=3 （重解）\lambda = 3 \text{ （\text{重解}）}

λ=3\lambda = 3 のとき

(−48−24)→(1−200)p1=C(21)\begin{pmatrix} -4 & 8 \\ -2 & 4 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & -2 \\ 0 & 0 \end{pmatrix} \quad p_1 = C\begin{pmatrix} 2 \\ 1 \end{pmatrix}

(2)

∣A−λE∣=∣−λ−132−λ∣=−λ(2−λ)+3=λ2−2λ+3λ=1±1−3=1±2i|A - \lambda E| = \begin{vmatrix} -\lambda & -1 \\ 3 & 2-\lambda \end{vmatrix} = -\lambda(2-\lambda) + 3 = \lambda^2 - 2\lambda + 3 \quad \lambda = 1 \pm \sqrt{1-3} = 1 \pm \sqrt{2}i

λ=1+2i\lambda = 1+\sqrt{2}i のとき

(−1−2i−131−2i)→((1+2i)(1−2i)1−2i31−2i)→(1+21−2i31−2i)→(31−2i31−2i)\begin{pmatrix} -1-\sqrt{2}i & -1 \\ 3 & 1-\sqrt{2}i \end{pmatrix} \rightarrow \begin{pmatrix} (1+\sqrt{2}i)(1-\sqrt{2}i) & 1-\sqrt{2}i \\ 3 & 1-\sqrt{2}i \end{pmatrix} \rightarrow \begin{pmatrix} 1+2 & 1-\sqrt{2}i \\ 3 & 1-\sqrt{2}i \end{pmatrix} \rightarrow \begin{pmatrix} 3 & 1-\sqrt{2}i \\ 3 & 1-\sqrt{2}i \end{pmatrix}

→(31−2i00)\rightarrow \begin{pmatrix} 3 & 1-\sqrt{2}i \\ 0 & 0 \end{pmatrix}

p1=C1(1−2i−3)=C111+2i((1−2i)(1+2i)−3(1+2i))=C111+2i(1+2−3(1+2i))p_1 = C_1 \begin{pmatrix} 1-\sqrt{2}i \\ -3 \end{pmatrix} = C_1 \frac{1}{1+\sqrt{2}i} \begin{pmatrix} (1-\sqrt{2}i)(1+\sqrt{2}i) \\ -3(1+\sqrt{2}i) \end{pmatrix} = C_1 \frac{1}{1+\sqrt{2}i} \begin{pmatrix} 1+2 \\ -3(1+\sqrt{2}i) \end{pmatrix}

=C131+2i(−1(1+2i))=C1(−11+2i)= C_1 \frac{3}{1+\sqrt{2}i} \begin{pmatrix} -1 \\ (1+\sqrt{2}i) \end{pmatrix} = C_1 \begin{pmatrix} -1 \\ 1+\sqrt{2}i \end{pmatrix}

λ=1−2i\lambda = 1-\sqrt{2}i のとき

(−1+2i−131+2i)→((1−2i)(1+2i)1+2i31+2i)→(1+21+2i31+2i)→(31+2i31+2i)\begin{pmatrix} -1+\sqrt{2}i & -1 \\ 3 & 1+\sqrt{2}i \end{pmatrix} \rightarrow \begin{pmatrix} (1-\sqrt{2}i)(1+\sqrt{2}i) & 1+\sqrt{2}i \\ 3 & 1+\sqrt{2}i \end{pmatrix} \rightarrow \begin{pmatrix} 1+2 & 1+\sqrt{2}i \\ 3 & 1+\sqrt{2}i \end{pmatrix} \rightarrow \begin{pmatrix} 3 & 1+\sqrt{2}i \\ 3 & 1+\sqrt{2}i \end{pmatrix}

→(31+2i00)\rightarrow \begin{pmatrix} 3 & 1+\sqrt{2}i \\ 0 & 0 \end{pmatrix}

p1=C2(1+2i−3)=C2⋅11−2i((1+2i)(1−2i)−3(1−2i))=C2⋅11−2i(1−2i−3(1−2i))p_1 = C_2\left(\frac{1+\sqrt{2i}}{-3}\right) = C_2 \cdot \frac{1}{1-\sqrt{2i}}\left(\frac{(1+\sqrt{2i})(1-\sqrt{2i})}{-3(1-\sqrt{2i})}\right) = C_2 \cdot \frac{1}{1-\sqrt{2i}}\left(\frac{1 - 2i}{-3(1-\sqrt{2i})}\right)

=C231−2i(−11−2i)=C2(−11−2i)= C_2 \frac{3}{1-\sqrt{2i}}\left(\frac{-1}{1-\sqrt{2i}}\right) = C_2\left(\frac{-1}{1-\sqrt{2i}}\right)

解説: r31bn1z