Extension problem 
Let $R$ be the ring consisting of the upper triangular $3\times3$ matrices over $\mathbb{Q}$ and $M_{1},\,M_{2}$ two $R$-modules defined as follows: for $A=(a_{ij})\in R$, $q\in \mathbb{Q}$   $$A\cdot q= a_{11}q,\,\,\,\,\,\,A\cdot q= a_{22} q$$
  I want to show that if $0\longrightarrow M_{1}\longrightarrow X\longrightarrow M_{2}\longrightarrow 0$ and  $0\longrightarrow M_{1}\longrightarrow Y\longrightarrow M_{2}\longrightarrow 0$ are two non split exact sequences, then $X$ is isomorphic to $Y$. 

I know that this is relevant to $Ext_{1}(M_{2}, M_{1})$, but i am very confused about that. Any ideas? thank you
 A: If $R$ is any ring and $1 = e_1 + ... + e_n$ is a decomposition of $1$ as the sum of pairwise orthogonal idempotents, then any $R$-module $M$ naturally decomposes, as an abelian group, as $M = \bigoplus\limits_{i=1}^{n} e_i M$. If $x\in R$ satisfies $x = e_i x = x e_j$, then the action of $x$ on $M$ annihilates all summands $M_k$ for $k\neq j$, and maps $M_j$ into $M_i$. In this way, $R$-modules can be viewed as $\{1,2,...,n\}$-indexed abelian groups $M = \{M_i\}$ together with maps $M_j\to M_i$ for each $x\in e_i R e_j$ (this is intentionally kept vague at this point).
In your example, $R = \tiny\begin{pmatrix} \ast & \ast & \ast \\ 0 & \ast & \ast \\ 0 & 0 & \ast\end{pmatrix}$ admits a decomposition $1 = e_1 + e_2 + e_3$ with $e_1 := \tiny\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0& 0 & 0 \end{pmatrix}$ etc., hence any $R$-module $M$ decomposes as $M = M_1\oplus M_2\oplus M_3$ with $M_i := e_i M$ - since $R$ is a ${\mathbb Q}$-algebra, this is even a decomposition as ${\mathbb Q}$-vector spaces. Moreover, the element $e_{12} := \tiny\begin{pmatrix} 0 & 1 & 0 \\  0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}$ and the analogously defined $e_{13}$ and $e_{23}$ act as ${\mathbb Q}$-linear homomorphisms $\alpha: M_2\to M_1$, $\gamma: M_3\to M_1$ and $\beta: M_3\to M_2$, respectively, and by associativity of the $R$-action, $e_{12} e_{23} = e_{13}$ implies that the composition $M_3\xrightarrow{\beta} M_2\xrightarrow{\alpha} M_1$ equals $M_3\xrightarrow{\gamma} M_1$.
All in all, this means that $R\text{-Mod}$ is equivalent to the category of diagrams of ${\mathbb Q}$-vector spaces of the form $\bullet\leftarrow\bullet\leftarrow\bullet$, with commutative diagrams as morphisms.
In this description, your module $M_2$ corresponds to $0\leftarrow {\mathbb Q}\leftarrow 0$, while $M_1$ corresponds to ${\mathbb Q}\leftarrow 0\leftarrow 0$. Now it is clear that the only extensions between $M_2$ and $M_1$ are of the form $E_\lambda :\equiv {\mathbb Q}\stackrel{\lambda}{\leftarrow}{\mathbb Q}\leftarrow 0$ with $\lambda\in{\mathbb Q}$, and they are non-split if and only if $\lambda\neq 0$. Moreover, $E_\lambda\cong E_\mu$ for $\lambda,\mu\in{\mathbb Q}\setminus\{0\}$ via multiplication by $\frac{\lambda}{\mu}$ in either the first or second copy of ${\mathbb Q}$.
