Prove: $\newcommand{\adj}{\operatorname{adj}}$If $A$ is invertible, then $\adj(A)$ is invertible and $[\adj(A)]^{-1}=\frac{1}{\det(A)}A=\adj(A^{-1})$

I can show the left side:

\begin{align*} A^{-1}&=\frac{1}{\det(A)}\adj(A)\\ \implies AA^{-1}&=\frac{1}{\det(A)}A \cdot \adj(A)\\ \implies I&=\frac{1}{\det(A)}A\cdot \adj(A), \end{align*} and, \begin{align*} A^{-1}A&=\adj(A)\frac{1}{\det(A)}A\\ \implies I&=\adj(A)\frac{1}{\det(A)}A. \end{align*} So, $$[\adj(A)]^{-1}=\frac{1}{\det(A)}A.$$

But I'm not sure how to show: $$\frac{1}{\det(A)}A=\adj(A^{-1}).$$

  • 3
    $\begingroup$ $$adj(A^{-1}) = \frac{1}{det(A^{-1})} A^{-1}$$ The left side gives you an expression of $A^{-1}$. Can you proceed then? $\endgroup$ – user27126 Mar 19 '14 at 3:06
  • $\begingroup$ I don't see where this is going? It says: $$adj(A^{-1})=\frac{1}{det(A)}A$$ $\endgroup$ – user46372819 Mar 19 '14 at 3:34

$$\newcommand{\adj}{\operatorname{adj}}A^{-1}=\frac{1}{\det(A)}\adj(A)$$ So,

\begin{align*} (A^{-1})^{-1}&=\frac{1}{\det(A^{-1})}\adj(A^{-1})\\ \iff A&=\frac{1}{\det(A^{-1})}\adj(A^{-1})\\ \iff \det(A^{-1})A &=\adj(A^{-1})\\ \iff \frac{1}{\det(A)}A&=\adj(A^{-1}). \end{align*}


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