If  $B=\left[\begin{array}{ccc}5& 2\alpha & 1\\ 0& 2& 1\\ \alpha & 3& -1\end{array}\right]$ is the inverse of a  $3\times 3$ matrix A, then the sum of all values of $\alpha$  for which $\left|A\right|+1=0$, is 

$\text{ Given matrix }B\text{ is the inverse matrix of }3\times 3\text{ matrix }A\text{, }$

$\text{ Where }B=\left[\begin{array}{ccc}5& 2\alpha & 1\\ 0& 2& 1\\ \alpha & 3& -1\end{array}\right]$

We know that ,

$\det \left(A\right)=\frac{1}{det\left(B\right)}\left[\because \det \left(A^{-1}\right)=\frac{1}{\det \left(A\right)}\right]$

Since, det(A) +1=0

$\begin{array}{l}\frac{1}{\det \left(B\right)}+1=0\\ \Rightarrow \det \left(B\right)=-1\\ \Rightarrow 5(-2-3)-2\alpha (0-\alpha )+1(0-2\alpha )=-1\\ \Rightarrow -25+2{\alpha }^2-2\alpha =-1\\ \Rightarrow 2{\alpha }^2-2\alpha -24=0\\ \Rightarrow {\alpha }^2-\alpha -12=0\\ \Rightarrow (\alpha -4)(\alpha +3)=0\end{array}$

$\Rightarrow \alpha =-3,4$

$\text{ So, required sum of all values of }\alpha \text{ is }4-3=1$