I have a matrix $A=\begin{bmatrix}r_{11}& r_{21} &r_{31}&r_{41}\\ r_{12}&r_{22}&r_{32}&r_{42}\\ r_{13}&r_{23}&r_{33}&r_{43}\end{bmatrix}$, As we see that taking all the three rows together we can get the combination $\{r_{i1}, r_{i2},r_{i3}\},~\forall i=1,2,3,4$, (in fact i need any one combination). Now what i want is, to transform this matrix in a deterministic manner so that the resultant matrix is a $4\times 3$ matrix $B$ such that taking any $3$ rows again gives me the set $\{r_{i1}, r_{i2},r_{i3}\}~\text{for some}~ i=1,2,3,4$, but no two rows gives me the combination. One such construction can be $B=\begin{bmatrix}r_{41}& r_{31} &r_{13}\\ r_{32}&r_{12}&r_{23}\\ r_{11}&r_{22}& r_{42}\\ r_{21}&r_{43} &r_{33}\end{bmatrix}$. As can be seen by combining rows $1,2,3$ we get the combination $r_{11},r_{12},r_{13}$, by combining rows $1,2,4$, we get the combination $r_{31},r_{32},r_{33}$,similarly the other rows. I have some questions, now
$1$.Since i have constructed a $4\times 3$ matrix that does what i want, how to prove this existence mathematically?
$2$.How many such arrangements $B$ can be made that guarentee the combination?
$3$ How to find a transformation matrix that converts the $3\times 4$ scheme to this $4\times 3$ scheme.
You already have: a constructive proof is a proof.
That depends on what symmetries you want to count. There are 4 sets of rows: you can assign those four sets to the four columns of $A$ in $4!$ ways; then you can permute each column, for a factor of $3!^4$, and you can permute each row of $B$, for a factor of $3!^4$, giving a total of $4!\;3!^8 = 4 \times 6^9 = 40310784$.
If you're looking for a matrix $C$ such that $B = CA$ then it can't be done, as indicated by the dimensions. If ${B = CA}^T$ is acceptable, that gives you a system of $16$ linear equations to solve in $12$ unknowns. There may be no solutions.