2D Matrix Transformation Invariance The transformation $\textbf{T}$ maps points $(x,y)$ of the plane into image points $(x', y')$ such that
$$\begin{align*} x' &= 4x + 2y + 14 \\ y' &= 2x + 7y + 42 \end{align*}$$
Find the coordinates of the invariant point of $\textbf{T}$. Hence express $\textbf{T}$ in the form $$\begin{pmatrix} x' \\ y' + k \end{pmatrix} = \textbf{A} \begin{pmatrix} x \\ y + k \end{pmatrix}$$ where $k$ is a positive integer and $\textbf{A}$ is a $2 \times 2$ matrix.

I found that the invariant point is $(0, -7)$ by setting $x = x'$ and $y = y'$. But I am a bit confused about what to do now. I tried some rearrangement to obtain $$\begin{pmatrix} x' \\ y' \end{pmatrix} + \begin{pmatrix} 0 \\ k \end{pmatrix} = \textbf{A} \begin{pmatrix} x \\ y \end{pmatrix} + \textbf{A}\begin{pmatrix} 0 \\ k \end{pmatrix}$$ Let's assume that $\textbf{A}$ is the matrix which represents the transformation $\textbf{T}$, then $$\begin{pmatrix} 0 \\ -7 \end{pmatrix} = \textbf{A} \begin{pmatrix} 0 \\ -7 \end{pmatrix} \implies \begin{pmatrix} 0 \\ 7 \end{pmatrix} = \textbf{A} \begin{pmatrix} 0 \\ 7 \end{pmatrix}$$ So from the earlier equation, $$\begin{pmatrix} x' \\ y' \end{pmatrix} = \textbf{A} \begin{pmatrix} x \\ y \end{pmatrix}$$ But the issue is that I cannot find any $\textbf{A}$ which represents the transformation. Can anyone please help me find $\textbf{A}$?

A transformation of the form $A\begin{bmatrix} x\\ y\end{bmatrix} $ is linear: it maps the origin to the origin; that's not what your $T$ does. For an affine transformation, you need to add the constant term. So your $T$ is of the form 
$$
T\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix} 4&2\\ 2&7\end{bmatrix} \begin{bmatrix} x\\ y\end{bmatrix}+ \begin{bmatrix} 14\\ 42\end{bmatrix}.
$$
So you have $$Tv=Av+r.$$A fixed point will satisfy $$Av+r=v,$$ which we may rewrite as $$(A-I)v=-r.$$ So that's the linear system you are looking for: 
$$
\begin{bmatrix} 4-1&2\\ 2&7-1\end{bmatrix} \begin{bmatrix} x\\ y\end{bmatrix}=-\begin{bmatrix} 14\\ 42\end{bmatrix}.
$$
