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 $pq$.
