\(2\) 次の正方行列 \(A = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)\) について以下の問いに答えよ. ただし \(a , b , c , d\) は実数とする.
(1) \(A^2 = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)\) を満たす \(A\) は存在しないことを示せ.
(2) \(A^2 = \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right)\) を満たす \(A\) をすべて求めよ.
(3) (2) で求めた \(A\) のそれぞれについて \(A+A^2+A^3+ \cdots +A^{2013}\) を求めよ.
【 解 答 】
(1)
条件をみたす行列 \(A\) が存在すると仮定すると
\[
A^2 = \left( \begin{array}{cc} a^2+bc & b(a+d) \\ c(a+d) & d^2+bc \end{array} \right) = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)
\]
なので
\[
\left\{\begin{array}{ll} a^2+bc = d^2+bc = 0 & ... [1] \\ b(a+d) = c(a+d) = 1 & ... [2] \end{array}\right.
\]
[2] より, \(a+d \neq 0\) ... [3] であり
\[
b = c \neq 0 \quad ... [4]
\]
これを [1] に代入すると
\[\begin{gather}
a^2 +b^2 = d^2 +b^2 = 0 \\
\text{∴} \quad a = d = b = c = 0
\end{gather}\]
これは, [3] [4] に矛盾する.
よって, \(A^2 = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)\) をみたす \(A\) は存在しない.
(2)
条件より \[ A^2 = \left( \begin{array}{cc} a^2+bc & b(a+d) \\ c(a+d) & d^2+bc \end{array} \right) = \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right) \] なので \[ \left\{\begin{array}{ll} a^2+bc = d^2+bc = 0 & ... [5] \\ b(a+d) = 1 & ... [6] \\ c(a+d) = -1 & ... [7] \end{array}\right. \] [6] [7] より, \(a+d \neq 0\) ... [8] で \[ b = -c = \dfrac{1}{a+d} \quad ... [9] \] [5] より \[\begin{align} a^2 & = d^2 \\ \text{∴} \quad d & = a \quad ( \ \text{∵} \ [8] \ ) \quad ... [10] \end{align}\] [9] [10] を [5] に代入すれば \[\begin{gather} a^2 -\left( \dfrac{1}{2a} \right)^2 = 0 \\ a^4 = \dfrac{1}{4} \\ \text{∴} \quad a = \pm \dfrac{1}{\sqrt{2}} \end{gather}\] よって, 求める行列 \(A\) は \[ A = \underline{\pm \dfrac{1}{\sqrt{2}} \left( \begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array} \right)} \]
(3)
\[\begin{align}
A^4 & = \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right) \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right) = -E \\
A^8 & = (-E)^2 = E
\end{align}\]
なので
\[\begin{align}
\textstyle\sum\limits _ {k=1}^8 A^k & = A+A^2+A^3-E-A-A^2-A^3+E \\
& = O
\end{align}\]
したがって, 自然数 \(n\) に対して
\[
\textstyle\sum\limits _ {k=n}^{n+8} A^k = O
\]
が成り立つ.
これを用いれば, \(2013 = 8 \cdot 251 + 5\) なので, 求める和 \(S\) は
\[
S = \textstyle\sum\limits _ {k=1}^5 A^k = A^2+A^3-E
\]
ここで
\[\begin{align}
A^3 & = \pm \dfrac{1}{\sqrt{2}} \left( \begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array} \right) \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right) \\
& = \pm \dfrac{1}{\sqrt{2}} \left( \begin{array}{cc} -1 & 1 \\ -1 & -1 \end{array} \right)
\end{align}\]
なので
\[\begin{align}
S & = \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right) \pm \dfrac{1}{\sqrt{2}} \left( \begin{array}{cc} -1 & 1 \\ -1 & -1 \end{array} \right) -\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) \\
& = \underline{\left( 1 \pm \dfrac{1}{\sqrt{2}} \right) \left( \begin{array}{cc} -1 & 1 \\ -1 & -1 \end{array} \right)}
\end{align}\]