Let $R$ be the set of all real numbers. Let $f: R \rightarrow R$ be a function defined by

$$ f(x)=\left\{\begin{array}{rcc} 2 x-5, & \text { if } & x<-3 \\ x+2, & \text { if } & -3 \leq x<5 \\ 3 x+1, & \text { if } & x \geq 5 \end{array}\right. $$

Match the following

$$ \begin{array}{llll} \hline & \text { List I } & & \text { List II } \\ \hline \text { A } & f(-5)+f(0)+f(-1)= & \text { I } & 16 \\ \hline \text { B } & f(f(5)+10 f(-3))= & \text { II } & 40 \\ \hline \text { C } & f(|f(-4)|)= & \text { III } & -32 \\ \hline \text { D } & f(f(f(1)))= & \text { IV } & -12 \\ \hline & & \text { V } & 19 \\ \hline \end{array} $$

If $A+B=\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2\end{array}\right], A B=\left[\begin{array}{lll}1 & 2 & 2 \\ 1 & 1 & 0 \\ 1 & 2 & 1\end{array}\right]$, then $A^2+B(A+B)=$

If $A, P, B$ are $3 \times 3$ matrices. If $|-B|=5,\left|B A^T\right|=15$, $\left|P^T A P\right|=-27$, then one of the values of $|P|$ is

If $A$ is a $3 \times 3$ matrix and $|A|=\frac{1}{2}$, then $\left|A^{-1}(\operatorname{Adj}(\operatorname{Adj} A))\right|^{-1}=$