Seja \mathbf{A}=\left[\begin{array}{cc}0 & 1 \\\\ -1 & 1\end{array}\right].
a) Calcule \mathbf{A}^{2}, \mathbf{A}^{3}, \mathbf{A}^{4}, \mathbf{A}^{5}, \mathbf{A}^{6}, \mathbf{A}^{7}.
b) Determine a matriz \mathbf{A}^{2001}.

Passo 1

O enunciado dá a matriz
\mathbf{A} = \begin{pmatrix}0 & 1 \\\\ -1 & 1\end{pmatrix}
e pede as potências \mathbf{A}^2,\dots,\mathbf{A}^7 e depois \mathbf{A}^{2001}. Vamos usar a definição de multiplicação de matrizes e observar que aparece uma periodicidade nas potências.

Passo 2

Calculando
\mathbf{A}^2 = \mathbf{A}\,\mathbf{A} = \begin{pmatrix}0 & 1 \\\\ -1 & 1\end{pmatrix} \begin{pmatrix}0 & 1 \\\\ -1 & 1\end{pmatrix} = \begin{pmatrix} 0\cdot0 + 1\cdot(-1) & 0\cdot1 + 1\cdot1 \\[6pt] -1\cdot0 + 1\cdot(-1) & -1\cdot1 + 1\cdot1 \end{pmatrix} = \begin{pmatrix}-1 & 1 \\\\ -1 & 0\end{pmatrix}.]

Passo 3

Agora
\mathbf{A}^3 = \mathbf{A}^2\,\mathbf{A} = \begin{pmatrix}-1 & 1 \\\\ -1 & 0\end{pmatrix} \begin{pmatrix}0 & 1 \\\\ -1 & 1\end{pmatrix} = \begin{pmatrix} -1\cdot0 + 1\cdot(-1) & -1\cdot1 + 1\cdot1 \\[6pt] -1\cdot0 + 0\cdot(-1) & -1\cdot1 + 0\cdot1 \end{pmatrix} = \begin{pmatrix}-1 & 0 \\\\ 0 & -1\end{pmatrix} = -\mathbf{I}.]

Passo 4

Sabendo que \mathbf{A}^3 = -\mathbf{I}, temos
\mathbf{A}^4 = \mathbf{A}^3\,\mathbf{A} = -\mathbf{I}\,\mathbf{A} = -\mathbf{A},\quad \mathbf{A}^5 = \mathbf{A}^4\,\mathbf{A} = -\mathbf{A}^2,\quad \mathbf{A}^6 = \mathbf{A}^5\,\mathbf{A} = -\mathbf{A}^3 = \mathbf{I},
\mathbf{A}^7 = \mathbf{A}^6\,\mathbf{A} = \mathbf{I}\,\mathbf{A} = \mathbf{A}.
Substituindo as matrizes já conhecidas:
\mathbf{A}^4 = -\begin{pmatrix}0 & 1 \\\\ -1 & 1\end{pmatrix} = \begin{pmatrix}0 & -1 \\\\ 1 & -1\end{pmatrix},\quad \mathbf{A}^5 = -\begin{pmatrix}-1 & 1 \\\\ -1 & 0\end{pmatrix} = \begin{pmatrix}1 & -1 \\\\ 1 & 0\end{pmatrix},
\mathbf{A}^6 = \begin{pmatrix}1 & 0 \\\\ 0 & 1\end{pmatrix},\quad \mathbf{A}^7 = \begin{pmatrix}0 & 1 \\\\ -1 & 1\end{pmatrix}.

Passo 5

Como \mathbf{A}^6 = \mathbf{I}, as potências de \mathbf{A} são periódicas de período 6. Para n=2001 calculamos o resto da divisão de 2001 por 6:
2001 \bmod 6 = 3,
logo
\mathbf{A}^{2001} = \mathbf{A}^3 = -\mathbf{I} = \begin{pmatrix}-1 & 0 \\[4pt] 0 & -1\end{pmatrix}.]

Resposta

\displaystyle \mathbf{A}^2 = \begin{pmatrix}-1 & 1 \\\\ -1 & 0\end{pmatrix} \displaystyle \mathbf{A}^3 = \begin{pmatrix}-1 & 0 \\\\ 0 & -1\end{pmatrix} \displaystyle \mathbf{A}^4 = \begin{pmatrix}0 & -1 \\\\ 1 & -1\end{pmatrix} \displaystyle \mathbf{A}^5 = \begin{pmatrix}1 & -1 \\\\ 1 & 0\end{pmatrix} \displaystyle \mathbf{A}^6 = \begin{pmatrix}1 & 0 \\\\ 0 & 1\end{pmatrix} \displaystyle \mathbf{A}^7 = \begin{pmatrix}0 & 1 \\\\ -1 & 1\end{pmatrix} \displaystyle \mathbf{A}^{2001} = \begin{pmatrix}-1 & 0 \\\\ 0 & -1\end{pmatrix}.

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