If $A=\left[\begin{array}{ccc}1& -2& 4\\ 2& -1& 3\\ 4& 2& 0\end{array}\right]$ is the adjoint of a square matrix B, then B^-1 is equal to:

Given: $\text{ adj }B=A$
therefore, $\det (\mathrm{adj}B)=\det \left(A\right)$
$\begin{array}{l}\Rightarrow (\det B{)}^2=\det A=-6+2(-12)+4(4+4)\\ \Rightarrow (\det B{)}^2=2\\ \Rightarrow \det B=\pm \sqrt{2}\\ \Rightarrow B^{-1}=\frac{1}{\det B}(\mathrm{adj}B)=\pm \frac{1}{\sqrt{2}}A\end{array}$