Step-by-step Solution:

\[ B=\begin{pmatrix} -1 & -1 & 2 \\ 0 & -1 & -1 \\ 0 & 0 & -1 \end{pmatrix}. \] \textbf{Step 1: Write } \(B=-I+N\). \[ B=-I+N, \quad N=\begin{pmatrix} 0 & -1 & 2 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{pmatrix}. \] Since \(N\) is strictly upper triangular, we have \(N^3=0\). \[ B^{19} = (-I+N)^{19} = \sum_{k=0}^{19} \binom{19}{k} (-I)^{19-k} N^k. \] Because \(N^3=0\), only terms with \(k=0,1,2\) survive: \[ B^{19} = (-I)^{19} + \binom{19}{1}(-I)^{18}N + \binom{19}{2}(-I)^{17}N^2. \] \textbf{Step 2: Simplify powers of \((-I)\).} \[ (-I)^{19} = (-1)^{19}I = -I, \quad (-I)^{18} = I, \quad (-I)^{17} = -I. \] So, \[ B^{19} = -I + 19N - \binom{19}{2}N^2. \] \textbf{Step 3: Compute \(N^2\).} \[ N=\begin{pmatrix} 0 & -1 & 2 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{pmatrix}, \quad N^2=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. \] \textbf{Step 4: Substitute.} \[ B^{19} = -I + 19\begin{pmatrix} 0 & -1 & 2 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{pmatrix} -171 \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. \] \[ B^{19} = \begin{pmatrix} -1 & -19 & 38 \\ 0 & -1 & -19 \\ 0 & 0 & -1 \end{pmatrix} + \begin{pmatrix} 0 & 0 & -171 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. \] \[ B^{19} = \begin{pmatrix} -1 & -19 & -133 \\ 0 & -1 & -19 \\ 0 & 0 & -1 \end{pmatrix}. \] \textbf{Step 5: Sum of all entries.} \[ S = (-1)+(-19)+(-133)+(-1)+(-19)+(-1) = -174. \] \[ \boxed{-174} \]