Let $A=\{f:[0,1]\to M_2(\mathbb{C}): $f continuous and $ f(0)=\begin{pmatrix} f_{11}(0) & 0 \\ 0 & f_{22}(0) \end{pmatrix}\}$ be a $C^*$-algebra with pointwise multiplication, involutions and norm $\|f\|=\sup\limits_{t\in [0,1]}\|f(t)\|_{op}$. I want to determine $\hat{A}=\{[\pi]:\pi$ is a irreducible representation of $A$\}, (Here is $\pi\sim \rho :\iff$ there is an unitary operator $V:H_{\pi}\to H_{\rho}$ such that $V\pi(a)=\rho(a)V$ for all $a\in A$). <br/>


I want to use the following fact: If $I$ is a closed ideal in a $C^*$-algebra $A$, it is $$\hat{A}=\hat{A}_I\coprod \hat{A}_{A/I}=\widehat{I}\coprod  \widehat{A/I},$$ where $\hat{A}_I=\{[\pi]\in \hat{A}: \pi(I)\neq 0\}$ and $\hat{A}_{A/I}=\{[\pi]\in \hat{A}: \pi(I)=0\}$. 

Consider $I=\{f\in A: f(0)=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\}$. $I$ is a closed Ideal in $A$, because $I$ is the kernel of the evaluation-homomorphism $\epsilon_0:A\to M_2(\mathbb{C}), f\mapsto f(0)$. Therefore $I$ is a $C^*$-subalgebra of $A$. Then $I$ is commuative and it follows that $I$ is isomorphic to $C_0((0,1],M_2(\mathbb{C}))$ as $C^*$-algebra. Maybe we get $\widehat{A/I}$ now, because we know what $A/I$ is, it is $A/I=A/ \ker(\epsilon_0)\cong Im(\epsilon_0)\cong \mathbb{C}^2$. But here I'm stuck, I'm not sure what $\widehat{A/I}$ should be and I really don't know what $\hat{I}$ is, I assume $\hat{I}$ is homeomorphic to $(0,1]$, but I can't prove it. 

I'm not sure if the question is ok for mathoverflow, I'm sorry if not.
But I appreciate your help.