Consider the matrix $$B=\begin{pmatrix}{-1} & {-1} & {2} \\ {0} & {-1} & {-1} \\ {0} & {0} & {-1}\end{pmatrix}$$. The sum of all the entries of the matrix $B^{19}$ is

The matrix is \[ B=\begin{pmatrix} -1 & -1 & 2\\ 0 & -1 & -1\\ 0 & 0 & -1 \end{pmatrix} \] Since \(B\) is upper–triangular with diagonal entries \(-1\), write \[ B = -I + N, \] where \[ N=\begin{pmatrix} 0 & -1 & 2\\ 0 & 0 & -1\\ 0 & 0 & 0 \end{pmatrix}, \qquad N^{3}=0. \] Using the binomial expansion: \[ B^{19}=(-I+N)^{19} = (-1)^{19}I + \binom{19}{1}(-1)^{18}N + \binom{19}{2}(-1)^{17}N^{2}. \] Compute signs: \[ (-1)^{19}=-1,\quad (-1)^{18}=1,\quad (-1)^{17}=-1. \] So: \[ B^{19} = -I + 19N - \binom{19}{2}N^{2}. \] Now compute sums: \[ \text{sum}(-I) = -3 \] \[ N^{2} = \begin{pmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix} \quad\Rightarrow\quad \text{sum}(N^{2}) = 1 \] \[ \Rightarrow -\binom{19}{2}\cdot 1 = -171 \] Final total: \[ -3 -171 = -174. \]