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Let
$$
A=left{fin C([0,1],M_{2}(mathbb{C})):f(1)=begin{pmatrix}xi & 0 \ 0 & lambdaend{pmatrix} text{ for some }xi,lambda text{ in } mathbb{C}right}.
$$
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.

We need to identify the irreducible representations of $A$. So assume that $pi:Ato B(H)$ 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 $x_0in [0,1]$ 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, $flongmapsto f(1)_{11}$ and $flongmapsto f(1)_{22}$.

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 $xin [0,1)cup{infty_1}cup{infty_2}$.

For the topology, let us write $hat t$ for the irrep $hat t(f)=f(t)$. Then, by definition of the Jacobson topology, $hat t_jtohat t$ if and only if $bigcap_{k}kerhat{t_{j_k}}subsetker t$ 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 $t_jto t$. 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 $infty_2$.