Find the tangent space of $\mathrm{Aff}(n)$ Find the tangent space of $\mathrm{Aff}(n)$. 
see Proof: Tangent space of the general linear group is the set of all squared matrices
$\mathrm{Aff}(n)$ is the set of all matrices of the form 
 $$     \begin{pmatrix}
        A & \ b \\
        0 & 1 \\
        \end{pmatrix}$$ where A $\in GL_n(\mathbb{R})$, b in $\mathbb{R}^n$, and $0$ is $1 \times n$ zero vector
Let   $$    1 =  \begin{pmatrix}
        I & \ 0 \\
        0 & 1 \\
        \end{pmatrix}$$
where I is identity matrix and 0 is a 0 column vector. 
So my path is Q(t)  $$     \begin{pmatrix}
        A(t) & \ bt \\
        0 & 1 \\
        \end{pmatrix}$$ where $A(t)$ is a path in $ GL_n(\mathbb{R})$ s.t $A(0) = I$. 
 A: Continuing from where you were:
$$\forall A:(-\epsilon,\epsilon) \to \operatorname{GL}_n(\mathbb R), A(0) = \operatorname{id}_n$$
$$\forall b \in \mathbb{R}^n$$ Define $Q :(-\epsilon,\epsilon) \to \operatorname{Aff}_n(\mathbb R)$ $$Q(t) =\begin{pmatrix}
        A(t) & \ bt \\
        0 & 1 \\
        \end{pmatrix}$$
Finding the tangent space at the identity means finding the tangent vectors to all possible curves at the identity. Thus, we differentiate and evaluate at $0$. $$\dot Q(t) =\begin{pmatrix}
        \dot A(t) & \ b \\
        0 & 0 \\
        \end{pmatrix}$$
 $$\dot Q(0) =\begin{pmatrix}
        \dot A(0) & \ b \\
        0 & 0 \\
        \end{pmatrix} \in \mathfrak{aff}(n) $$ 
$$\implies \mathfrak{aff}(n) =  \left\{ \begin{pmatrix}
        M & \ b \\
        0 & 0 \\
        \end{pmatrix} \ \middle|\ M \in \operatorname{Mat}_n(\mathbb R),b \in \mathbb{R}^n \right\}$$
Thus we find that $b$ is arbitrary, and varying $A$ over the general linear group gives us all possible matrices in $\mathfrak{gl}_n(\mathbb R) = \operatorname{Mat}_n(\mathbb R)$. 
Then as vector spaces, $\mathfrak{aff}_n(\mathbb R) = \mathfrak{gl}_n(\mathbb R) \oplus \mathbb R ^n $. As Lie algebras, you have to take the semi-direct product of the two terms, which in this case is just a way of saying the direct sum of the two as vector spaces has the bracket operation defined on it in the natural way.
