For $A=\begin{bmatrix}2&1\\1&1\end{bmatrix}$, $\det A=2-1=1$. The $2\times2$ inverse formula gives $A^{-1}=\begin{bmatrix}1&-1\\-1&2\end{bmatrix}$. Direct multiplication gives $AA^{-1}=I$, providing a quick check.
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