This is exercise IX.1 from "C* Algebras by Example" by Davidson.
Suppose that $(\mathcal{E}_{i},\varphi_{i})$ represent two extensions of $\mathcal{K}$ (compact operators) by $C(X)$. Show that the sum of extensions is given by
$$ \{\begin{bmatrix} E_{11} & K_{12}\\ K_{21} & E_{22} \end{bmatrix} : E_{ii} \in \mathcal{E}_{i}, K_{ij} \in \mathcal{K}, \varphi_{1}(E_{11})=\varphi_{2}(E_{22})\}$$ Could anyone give me a hint on how to approach this one?
The addition is defined for monomorphisms, $[\tau_1]+[\tau_2]=[\tau_1\oplus\tau_2]$. The exercise seems to ignore the identification between $M_2(B(H))$ and $B(H)$, and just wants the result on $M_2(B(H))$. $\def\E{\mathcal E}$
We will have $\E_1+\E_2=\pi_2^{-1}((\tau_1\oplus \tau_2)(C(X))$. In this context $\pi_2:M_2(B(H))\to M_2(B(H)/K(H))$ is the quotient map, which can be seen as the $2\times 2$ amplification of the usual quotient map. So we are looking at $$ \begin{bmatrix} \tau_1(f)&0\\0&\tau_2(f)\end{bmatrix}= \begin{bmatrix} \pi(T_{11})& \pi(T_{12})\\ \pi(T_{21})&\pi(T_{22})\end{bmatrix}. $$ This forces $\pi(T_{12})=\pi(T_{21})=0$, so $T_{12},T_{21}\in K(H)$, and $$ \varphi_1(T_{11})=\tau_1^{-1}\pi(T_{11})=f=\tau_2^{-1}\pi(f)=\varphi_2(T_{22}). $$