adjoint-operatorsdeterminantlinear algebra
Let $A$ be a $3\times3$ matrix such that $$\mathrm{adj}(A) = \begin{pmatrix}3 & -12 & -1 \\ 0 & 3 & 0 \\ -3 & -12 & 2\end{pmatrix}.$$Find the value of $\det(A)$.
I know that the adjoint of A is the transpose of the cofactors of $A$, but how do I find A or $\det(A)$ in reverse for a particular matrix? all I can think of is using the $A^{-1} = \det(A)/\mathrm{adj}(A)$ but I have no idea how to apply this for a direct value question.
Here is the answer from the solution paper.
If $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ then $\operatorname{adj} A = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. So you need to find $a,b,c,d$ such that $$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}. $$ In other words, you have to solve the system of linear equations $$ \begin{align*} a &= d, \\ b &= -b, \\ c &= -c, \\ d &= a. \end{align*} $$ You take it from here.
Best Answer
$$A^{-1} = \dfrac{\text{adj} A}{\det A} \Rightarrow\\ \det(A^{-1}) = \dfrac{\det(\text{adj} A)}{(\det A)^n} \Rightarrow\\ \dfrac{1}{\det A } = \dfrac{\det(\text{adj} A)}{(\det A)^n} \Rightarrow\\ (\det A)^{n-1} = \det(\text{adj} A)\Rightarrow\\ \large\boxed{\det A = \sqrt[\leftroot{-3}\uproot{1}n-1]{\det (\text{adj}A)}}$$