If $\frac{x^2+7}{\left(x^2+1\right)\left(x-2\right)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+1}$ 
then the determinant of the matrix $\left(\begin{array}{ll}A& B\\ C& \frac{2}{5}\end{array}\right)$ is

Given,

$\frac{x^2+7}{\left(x^2+1\right)\left(x-2\right)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+1}$

$\Rightarrow x^2+7=A\left(x^2+1\right)+\left(Bx+C\right)\left(x-2\right) ....\left(1\right)$

On taking $x=2$ we get,

$\Rightarrow 11=5A\Rightarrow A=\frac{11}{5}$

Now comparing coefficient of $x^2$ in equation $\left(1\right)$ we get,

$\Rightarrow 1=A+B\Rightarrow B=1-\frac{11}{5}\Rightarrow B=\frac{-6}{5}$

Now comparing constant term in equation $\left(1\right)$ we get,

$\Rightarrow 7=A-2C\Rightarrow 2C=\frac{11}{5}-7\Rightarrow C=\frac{-12}{5}$

Now putting the value in $\left|\begin{array}{ll}A& B\\ C& \frac{2}{5}\end{array}\right|=\left|\begin{array}{cc}\frac{11}{5}& \frac{-6}{5}\\ \frac{-12}{5}& \frac{2}{5}\end{array}\right|$

$\Rightarrow \left|\begin{array}{ll}A& B\\ C& \frac{2}{5}\end{array}\right|=\frac{11}{5}\times \frac{2}{5}-\left(\frac{-6}{5}\right)\times \left(\frac{-12}{5}\right)=\frac{22}{25}-\frac{72}{25}=\frac{-50}{25}=-2$