Think about $\left({\phantom-d\phantom--b\atop-c\phantom{--}a}\right)$ as $tI - A$ where $t=a+d$ is the trace of $A$. Since $A$ satisfies its own characteristic equation (Cayley-Hamilton), we have $A^2-tA+\Delta A^2 - t A + \Delta \cdot I = 0$ where $\Delta = ad-bc$ is the determinant. Thus $\Delta \cdot I = t A - A^2$. Now divide both sides by $\Delta \cdot A$ to get $A^{-1} = \Delta^{-1}(tI-A)$, QED.
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2 | Made the invocation of Cayley-Hamilton explicit, and linked it to Wikipedia. | ||
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Think about $\left({\phantom-d\phantom--b\atop-c\phantom{--}a}\right)$ as $tI - A$ where $t=a+d$ is the trace of $A$. Since $A$ satisfies its own characteristic equation, we have $A^2-tA+\Delta \cdot I = 0$ where $\Delta = ad-bc$ is the determinant. Thus $\Delta I = t A - A^2$. Now divide both sides by $\Delta \cdot A$ to get $A^{-1} = \Delta^{-1}(tI-A)$, QED. |
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