How to transform affine dynamical system into linear system of difference equations I have a linear difference equation for a system with the form
$c_j(t+1) = \frac{b}{n} + (1-b)\sum_{i=1}^{n}B_{ij}c_i(t)$
where B is a square matrix
I'm trying to get this into the form $c(t+1) = Ac(t)$ , where $A$ is a square matrix that governs the dynamics of the system. However, I'm having trouble because of the $\frac{b}{n}$ term.
I have:
$$\left( \begin{array}{}
c_1(t+1)\\
c_2(t+1) \\
 ...\\ 
c_n(t+1)\end{array} \right) =\left( \begin{array}{}
\frac{b}{n} + (1-b)(B_{11}c_1(t) + B_{21}c_2(t) + ... + B_{n1}c_n(t))\\
\frac{b}{n} + (1-b)(B_{12}c_1(t) + B_{22}c_2(t) + ... + B_{n2}c_n(t)) \\
 ...\\ 
 \frac{b}{n} + (1-b)(B_{1n}c_1(t) + B_{2n}c_2(t) + ... + B_{nn}c_n(t))\end{array} \right)$$
I want this in the form $c(t+1) = Ac(t)$
so that I can solve the system using $c(t) = A^tc(0) $, but I'm not seeing how to achieve this form. It may be that I'm approaching this entirely incorrectly.
 If you have thoughts on how to proceed, I would very much appreciate them.
edit: I would like it specifically in the form stated above without any additions to the c vector as suggested in the answers below. If Is this even possible?

Hint: Each row is multiplied by (1-b) except for the first term, we can therefore factor 1-b out to get:
$$\left( \begin{array}{c}
c_1(t+1)\\
c_2(t+1) \\
\vdots\\ 
c_n(t+1)\end{array} \right) =  (1-b)\left( \begin{array}{c}
\frac{b}{n(b-1)} + (B_{11}c_1(t) + B_{21}c_2(t) + ... + B_{n1}c_n(t))\\
\frac{b}{n(b-1)} + (B_{12}c_1(t) + B_{22}c_2(t) + ... + B_{n2}c_n(t)) \\
 \vdots\\ 
 \frac{b}{n(b-1)} + (B_{1n}c_1(t) + B_{2n}c_2(t) + ... + B_{nn}c_n(t))\end{array} \right)$$
Now let $\cases{c_0(t) = 1\\ B_{0k} = \frac{b}{n(b-1)}}$ and what happens then?
