Can $A^{100}B$ where ($A, B$ are matrices) be calculated provided $A$ is deficient? I have seen a procedure to calculate $A^{100}B$ like products without actually multiplying where $A$ and $B$ are matrices. But the procedure will work only if $A$ is diagonalizable because the procedure attempts to find $B$ such that
$B = a_1X_1 + a_2X_2 + ...$, where $X_1,X_2,...$ are independent eigen-vectors (or basis) of $A$ and $a_1,a_2,...$ are scalars. 
Is there any other procedure to multiple such matrices where there is no restriction on matrix $A$ being deficient or diagnalizable?
 A: When a matrix is not diagonalizable, you can instead use Jordan Normal Form (JNF).  Instead of picking a basis of eigenvectors, you use "approximate eigenvectors." While eigenvectors are things in the kernel of $A-\lambda I$, approximate eigenvectors are things in the kernel of $(A-\lambda I)^k$.  Essentially, we can reduce the problem to when $A$ is a single Jordan block.
Suppose that $A=\lambda I + N$ where $N$ is nilpotent, and let $x$ be a cyclic vector for $N$, so that our vector space has a basis of $x, Nx, N^2 x, \ldots, N^{k-1} x$.  For simplicity of notation, set $x_i=N^i x$.  This is essentially an abstract form of what it means to be a Jordan block, as $A$ is a put into Jordan form when we take the $x_i$ as a basis.
To mimic what you had for diagonalizable matrices, We need to compute $A^n x_i$.  Since $N$ commutes with $\lambda I$, we can use the binomial theorem to compute $(\lambda I + N)^n = \sum \binom{n}{i} \lambda^{n-i} N^i$.  Then 
$$(\lambda I + N)^n x_j = \sum \binom{n}{i} \lambda^{n-i} N^i x_j=\sum \binom{n}{i} \lambda^{n-i} x_{j+i}.$$
A: I think the fastest way in general is trying to calculate $A^{100}$ faster.
Have you heard of fast exponentiation algorithm?
When you have to calculate $A^{100}$, 
You can calculate $A^2, A^4, A^8....A^{64}$ and compute $A^{64} A^{32} A^{4}$ instead of multipling A 100 times. This will take about 10 matrix multiplication. 
Then, you can use $A^{100}$ to get the result what you want. 
