Rigorous Computation of the Spectrum of a Certain C$^{*}$-algebra Let 
$$
A=\left\{f\in C([0,1],M_{2}(\mathbb{C})):f(1)=\begin{pmatrix}\xi & 0 \\ 0 & \lambda\end{pmatrix} \text{ for some }\xi,\lambda \text{ in } \mathbb{C}\right\}.
$$
In his Operator Algebras book, Blackadar mentions that the spectrum of $A$ is $[0,1)\cup\{\infty_{1}\}\cup\{\infty_{2}\}$, where $\infty_{1}$ and $\infty_{2}$ are two distinct endpoints that go on the right-side of the interval. It is clear, then, that the spectrum of $A$ is not Hausdorff since $\infty_{1}$ and $\infty_{2}$ cannot be separated. 
This is intuitively clear to me; for I know that the spectrum of $C([0,1],M_{2}(\mathbb{C}))$ is $[0,1]$, and I am guessing that the two distinct right-hand points, $\infty_{1}$ and $\infty_{2}$, correspond to the two irreducible representations coming from the point evaluations at $1$. 

I was wondering if there is a simple, but rigorous, computation that
  justifies this calculation of $\hat{A}$ and its topology.

 A: We need to identify the irreducible representations of $A$. So assume that $\pi:A\to B(H)$ is irreducible. 
Note that the constant function $E_{11}$ is in $A$. Then $\pi(E_{11})$ is a projection in $B(H)$. We have 
$$
(\pi(E_{11})\pi(A)\pi(E_{11}))'=\pi(E_{11})\pi(A)'=\mathbb C\,\pi(E_{11}). 
$$
So the restriction $\pi:E_{11}AE_{11}\to \pi(E_{11})B(H)\pi(E_{11})$ is irreducible.
It is easy to check that $E_{11}AE_{11}\simeq C[0,1]$ (simply, the 1,1 coordinate is continuous). Now the irreducible representations of $C[0,1]$ are precisely the point evaluations. So there exists $x_0\in [0,1]$ such that $$\pi(E_{11} f E_{11})=f(x_0)_{11}\,\pi(E_{11}).$$
Assume first that $x_0<1$. By abuse of notation, I will denote by $E_{12}$ a function that is equal to $E_{12}$ on $(0,1-\delta)$, and decreases to $0$ on $(1-\delta,1)$; and we let  $E_{21}=E_{12}^*$. So all the equalities below hold on $(0,1-\delta)$; there is no issue because we may assume that $x_0<1-\delta$. 
For any $k,j=1,2$,
$$
\pi(E_{1k}fE_{j1})=\pi(E_{11}E_{1k}fE_{j1}E_{11})=(E_{1k}fE_{j1})(x_0)_{11}\,\pi(E_{11})=f(x_0)_{kj}\,\pi(E_{11}).
$$
Now 
\begin{align}
\pi(f_{kj}E_{kj})&=\pi(E_{k1}E_{1k}fE_{j1}E_{1j})=\pi(E_{kj})\pi(E_{1k}fE_{j1})\pi(E_{1j})\\
&=f(x_0)_{kj} \,\pi(E_{k1}E_{11}E_{1j})
=f(x_0)_{kj}\,\pi(E_{kj}).
\end{align}
Adding over $k,j$
$$
\pi(f)=\sum_{k,j} f(x_0)_{kj},\pi(E_{kj}).
$$
In particular, $\pi(A)$ is contained in the span of $\{\pi(E_{kj})\}_{k,j}$, and so $\dim H=2$. For different $\delta$ we'll get different $\pi(E_{12})$, but it will always be a partial isometry between $\pi(E_{11})$ and $\pi(E_{22})$ (and these two do not depend on $\delta$). 
When $x_0=1$, the above doesn't work because $E_{12}(1)=0$ and we don't have the partial isometries we need. So we need to consider these two separate irreducible representations, $f\longmapsto f(1)_{11}$ and $f\longmapsto f(1)_{22}$. 
Thus the spectrum is what it needs to be, $[0,1)\cup\{\infty_1\}\cup\{\infty_2\}$. The primitive ideals are $K_x=\{f:\ f(x)=0\}$, for $x\in [0,1)\cup\{\infty_1\}\cup\{\infty_2\}$.
For the topology, let us write $\hat t$ for the irrep $\hat t(f)=f(t)$. Then, by definition of the Jacobson topology,  $\hat t_j\to\hat t$ if and only if $\bigcap_{k}\ker\hat{t_{j_k}}\subset\ker t$ for all subnets $\{\hat{t_{j_k}}\}$ of $\{\hat{t_j}\}$. This is precisely that $f(t_{k_j})=0$ for all $k$ implies that $f(t)=0$; as continuous functions separate points, this is saying that $t_j\to t$. So the topology on the spectrum of $A$ is the usual of $[0,1]$, with the exception that we have two "$1$", that we denoted $\infty_1$ and $\infty_2$. 
