Let $$A=\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix}$$ and B be two matrices such that $$A^{100}=100B+I$$. Then the sum of all the elements of $$B^{100}$$ is_______

We are given $$A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$$ and $$A^{100} = 100B + I$$. We need the sum of all elements of $$B^{100}$$.

Characteristic equation: $$\det(A - \lambda I) = (3 - \lambda)(-1 - \lambda) + 4 = \lambda^2 - 2\lambda + 1 = (\lambda - 1)^2 = 0$$.

By Cayley-Hamilton: $$(A - I)^2 = 0$$.

Let $$N = A - I = \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix}$$. Then $$N^2 = 0$$ (nilpotent).

(since $$N^2 = 0$$, all higher terms vanish).

Given $$A^{100} = 100B + I$$:

Since $$B = N$$ and $$N^2 = 0$$, we have $$B^2 = 0$$, and therefore $$B^{100} = (B^2)^{50} = 0^{50} = 0$$.

So $$B^{100}$$ is the zero matrix, and the sum of all elements is $$0$$.

The answer is $$\boxed{0}$$.