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Given $\mathrm{adj}(A)$ where $A$ is an $n\times n$ matrix, how do you find the value of $\det(A)$ and $A^{-1}$?

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    $\begingroup$ $A • adj(A)= det(A) I_n$. $\endgroup$ – Rubertos Jul 7 '15 at 22:18
  • $\begingroup$ You cannot find $\det A$ from $\operatorname{adj} A$ when $n = 1$, because the adjoint of any $1 \times 1$-matrix is the $1 \times 1$-matrix $\left(\begin{array}{cc} 1 \end{array}\right)$. For $n = 2$, it is easy. For greater $n$, I think you can reconstruct $A$ "up to an $n-1$-th root of unity" (and no better, because if $\zeta$ is an $n-1$-th root of unity, then $\zeta A$ and $A$ have the same adjoint). See my comment on math.stackexchange.com/questions/1353149/… for a first step. $\endgroup$ – darij grinberg Jul 7 '15 at 22:27
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Hint $$ \left( \begin{array}{ccc} \text{det}(A) & & 0\\ & \ddots & \\ 0 & & \text{det}(A) \end{array} \right)=\text{det}(A) \cdot \text{I}_n = A \cdot \text{adj}(A) $$ and $$ A^{-1}= \frac{1}{\text{det}(A)}\text{adj}(A) $$

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  • $\begingroup$ If A is 4x4 matrix how would you find the value of det(A) from adj(A). $\endgroup$ – SimpleCakes Jul 7 '15 at 22:23
  • $\begingroup$ Could I just add up one of the row or column of adj(A) to get det(A) $\endgroup$ – SimpleCakes Jul 7 '15 at 22:29
  • $\begingroup$ @SimpleCakes, Since $$ \text{det}(A) \cdot \text{I}_n = A \cdot \text{adj}(A) $$ Then $(\text{det}(A))^n= \text{det}(A)\cdot \text{det} ( \text{adj}(A)) $. Now obtain $\text{det}(A)$ from last equation in terms of $ \text{det} ( \text{adj}(A)) $ $\endgroup$ – Alonso Delfín Jul 7 '15 at 22:35

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