Given the adjoint $\mathrm{adj}(A)$, how do you find $\det(A)$ and $A^{-1}$?

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

• $A • adj(A)= det(A) I_n$. – Rubertos Jul 7 '15 at 22:18
• 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. – darij grinberg Jul 7 '15 at 22:27

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 $$A^{-1}= \frac{1}{\text{det}(A)}\text{adj}(A)$$
• @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 $\text{det} ( \text{adj}(A))$ – Alonso Delfín Jul 7 '15 at 22:35