# 2

Letâ€™s take the questionâ€™s hint and try to identify a recurrence relation.

\mathbf{A}=\begin{pmatrix}-5&4\\-9&7\end{pmatrix}

\mathbf{A}^2=\begin{pmatrix}-5&4\\-9&7\end{pmatrix}\cdot\begin{pmatrix}-5&4\\-9&7\end{pmatrix}=\begin{pmatrix}-11&8\\-18&13\end{pmatrix}

\mathbf{A}^3=\begin{pmatrix}-5&4\\-9&7\end{pmatrix}\cdot\begin{pmatrix}-11&8\\-18&13\end{pmatrix}=\begin{pmatrix}-17&12\\-27&19\end{pmatrix}

\mathbf{A}^4=\begin{pmatrix}-5&4\\-9&7\end{pmatrix}\cdot\begin{pmatrix}-17&12\\-27&19\end{pmatrix}=\begin{pmatrix}-23&16\\-36&25\end{pmatrix}

This should be enough examples to see that: \mathbf{A}^{n+1}-\mathbf{A}^n=\begin{pmatrix}-6&4\\-9&6\end{pmatrix}

Letâ€™s prove this result by induction to confirm its validity for all n\ge1.
Base case: \mathbf{A}^2-\mathbf{A}^1=\begin{pmatrix}-6&4\\-9&6\end{pmatrix}
Assume that: \mathbf{A}^{k+1}-\mathbf{A}^k=\begin{pmatrix}-6&4\\-9&6\end{pmatrix}
Then:

\begin{align}\mathbf{A}^{k+2}-\mathbf{A}^{k+1}&=\mathbf{A}(\mathbf{A}^{k+1}-\mathbf{A}^k)\\ &=\begin{pmatrix}-5&4\\-9&7\end{pmatrix}\cdot\begin{pmatrix}-6&4\\-9&6\end{pmatrix}\\ &=\begin{pmatrix}-6&4\\-9&6\end{pmatrix} \end{align}

Hence, the result is true for all n\ge1.
From this result we can deduce that

\begin{align}\mathbf{A}^{n}&=\begin{pmatrix}-5-6(n-1)&4n\\-9n&7+6(n-1)\end{pmatrix}\\ &=\begin{pmatrix}-6n+1&4n\\-9n&6n+1\end{pmatrix}\\ \\ \implies \mathbf{A}^{1000} &=\begin{pmatrix}-6(1000)+1&4(1000)\\-9(1000)&6(1000)+1\end{pmatrix}\\ &=\underline{\begin{pmatrix}\mathbf{-5999}&\mathbf{4000}\\\mathbf{-9000}&\mathbf{6001}\end{pmatrix}} \end{align}

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Fantastic!!