When we translate a point $p_3 = (x,y,z)$ to coordinates $p_4 =(x + t_x , y + t_y ,z + t_z,1)$ we use $4 \times 4$ Translation matrix using homogenous coordinates, hence we add a $1$ to fourth coordinate of $p_4$.
But I could not find a matrix $M : R^3 \rightarrow R^4$ for which $M p_3 = p_4$. Note that I am talking about a $1$ on the fourt coordinate what ever vector $p_3$ be.
Why can not we find such a generic translation matrix. Omit the case of matrix addition i.e. $M$ may only be multiplication of other matrices.
Best Answer
You cannot represent a translation in $3$-dimensional space with a matrix smaller than $4\times 4$. All $3\times 3$ matrices represent transformations which leave the origin fixed, because multiplying a matrix by the zero vector always yields the zero vector. Homogeneous coordinates might be used as a trick to solve this, but you need to add a vector entry and so switch to a $4\times 4$ matrix.
ADDED LATER (taken from comments)
In comments, the OP asked for a coordinate-free explanation. This is done by distinguishing "linear" vs "affine" or, more generally, "projective" transformations. Given a vector space $V$, one can build its projective completion $\mathbb P(V\oplus \mathbb K)=\left[V\oplus \mathbb K \setminus\{(0,0)\}\right]/ \sim$, where $(v, \lambda)\sim(w, \mu)$ if and only if $ A(v, \lambda)=B(w, \mu)$ for scalars $A, B\in\mathbb K$. The space $V$ naturally embeds into its projective completion via the map $v\mapsto (v, 1)$. (This is the coordinate-free realization of the introduction of homogeneous coordinates on $\mathbb K^n$).
This completion allows one to algebraically represent more transformations of $V$ than just the linear ones. The keyword here is projective transformation. Translations are realized as projective transformation via the following trick: \begin{equation} \begin{array}{ccc} T_h(v)=v+h & \leftrightarrow&\begin{bmatrix} I_V & h \\ 0 & 1\end{bmatrix} \begin{bmatrix} v \\ 1\end{bmatrix} = \begin{bmatrix} v+h \\ 1\end{bmatrix} \end{array} \end{equation} The relevance of this to the question is that the translation is realized as a transformation of the projective completion of $V$. This needs one more coordinate to be described as a matrix, hence (if $\dim V=3$) a $4\times 4$ matrix necessarily.