Let $A=\left(\begin{array}{cc}4& -2\\ \alpha & \beta \end{array}\right)$. If $A^2+\gamma A+18I=O$, then $det\left(A\right)$ is equal to _______.

Given $A=\left[\begin{array}{cc}4& -2\\ \alpha & \beta \end{array}\right] \Rightarrow \det \left(A\right)=4\beta +2\alpha$

Characteristic equation of the matrix is 

$\left[\begin{array}{cc}4-\lambda & -2\\ \alpha & \beta -\lambda \end{array}\right]=0$

$\Rightarrow 4\beta +{\lambda }^2-\left(\beta +4\right)\lambda +2\alpha =0$

$\Rightarrow {\lambda }^2-\left(\beta +4\right)\lambda +2\alpha +4\beta =0$

Comparing with given equation $A^2+\gamma A+18I=0$, we get

$\Rightarrow \gamma =-\left(\beta +4\right) \& 2\alpha +4\beta =18$

Hence $det\left(A\right)=4\beta +2\alpha =18$