Projective transformation for ideal points

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I know that for a projective transformation $T:PR^2 \mapsto PR^2$ it will have the form of $$T(X)=AX$$ where A is a matrix and X is a vector in homogeneous coordinates.

I also know that this $T$ maps ideal points to itself. $ie,$
$$T(X)=X$$

So $X$ represented in homogeneous coordinates will be $\begin{pmatrix} x\\ y\\ 0\\ \end{pmatrix}$

then if I were to represent the mapping of $X$, it will look like the following $$T(X)= AX= \begin{pmatrix} a_1&b_1&c_1\\ a_2&b_2&c_3\\ a_3&b_3&c_3\\ \end{pmatrix} * \begin{pmatrix} x\\ y\\ 0\\ \end{pmatrix} = \begin{pmatrix} x\\ y\\ 0\\ \end{pmatrix} $$

How would I prove that this transformation is in fact a translation or radial transformation.

I have tried multiplying the matrix and the vector and I got something like this: $$a_1x+b_1y+0=x\\ a_2x+b_2y+0=y\\ a_3x+b_3y+0=0\\ $$ from here I have trouble showing that it is a translation or radial transformation