# SL Decomposition of $\begin{pmatrix} 5 & -4 \\ 9 & -7 \end{pmatrix}$

I want to find a decomposition of this matrix in the form of $$A^{n_1} \cdot B \cdot A^{n_2} \cdot B \cdots B \cdot A^{n_k}$$ where $$A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$ I am aware that $A^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$, but fail at using the Euclidean algorithm as in Example 2.1 to find a decomposition.

I tried using results from here, but that didn´t help me, either.

$$B \mapsto \left( \begin{array}{cc} 9&-7 \\ -5&4 \\ \end{array} \right)$$ $$A^2 \mapsto \left( \begin{array}{cc} -1&1 \\ -5&4 \\ \end{array} \right)$$ $$B \mapsto \left( \begin{array}{cc} -5&4 \\ 1&-1 \\ \end{array} \right)$$ $$A^{5} \mapsto \left( \begin{array}{cc} 0&-1 \\ 1&-1 \\ \end{array} \right)$$ $$B \mapsto \left( \begin{array}{cc} 1&-1 \\ 0&1 \\ \end{array} \right)$$ $$A \mapsto \left( \begin{array}{cc} 1&0 \\ 0&1 \\ \end{array} \right)$$
• So the answer ist $B^{-1} A^{-2} B^{-1} A^{-5} B^{-1} A^{-1}$. – Viktor Glombik May 20 '18 at 16:02