4
$\begingroup$

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: $$\frac{1}{\det(A)}A=\adj(A^{-1}).$$

$\endgroup$
2
  • 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
5
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.