The sum of all four entries of the matrix A^-50 is

Write $A=I+N$ with $N=\begin{pmatrix}2&-1\\4&-2\end{pmatrix}$. Then $N^2=\begin{pmatrix}4-4&-2+2\\8-8&-4+4\end{pmatrix}=O$, so $N$ is nilpotent. Hence, by the binomial theorem (terms with $N^2$ and higher vanish and $I$, $N$ commute), $A^{k}=I+kN$ for every integer $k$, including negatives, because $A^{-1}=I-N$ satisfies $(I+N)(I-N)=I$. Therefore $A^{-50}=I-50N=\begin{pmatrix}-99&50\\-200&101\end{pmatrix}$, whose entries sum to $-99+50-200+101=-148$. The value $152$ is the sum for $A^{50}=I+50N$ and is the common mistake of ignoring the negative exponent. Nilpotent decomposition is far faster than diagonalising here because the only eigenvalue is 1 with a single Jordan block.