# How Find the value $\det{(A)}$ if know $A\cdot\begin{bmatrix}1 \end{bmatrix}$

let $A_{n\times n}$matrix,and $A^{*}$is Adjugate matrix of the $A$,$p,q>0$ is give numbers,and such following condition $$A\cdot\begin{bmatrix} 1\\ 1\\ \vdots\\ 1 \end{bmatrix}=\begin{bmatrix} p\\ p\\ \vdots\\ p \end{bmatrix},and ,A^{*}\cdot\begin{bmatrix} 1\\ 1\\ \vdots\\ 1 \end{bmatrix}=\begin{bmatrix} q\\ q\\ \vdots\\ q \end{bmatrix}$$ and $A^{-1}$ is exsit,Find the $\det{(A)}$

My idea: I knw this matrix $A$ one eigenvalue is $p$,and $A^{*}$ have one eigenvalue is $q$ so $$AX=pX,A^{*}Y=qY$$

There is a theorem which states that $$A \cdot A^*= A^* \cdot A= \det(A) \cdot I$$ Where $I$ is the identity matrix.
So, applying $A \cdot A^*$ to the vector $[1, \dots, 1]^T$ you get that the determinant of $A$ is $pq$.