As is known (see Kadison-Ringrose, 3.4.1) each closed ideal $I$ in the $C^*$-algebra $C(X)$ of continuous functions on a compact space $X$ has the form $$ I=\{f\in C(X): \ \forall x\in S\quad f(x)=0 \} $$ for some closed subset $S$ in $X$.
Is the same true for the two-sided ideals in the $C^*$-algebra $C(X,M_n)$ of continuous maps $f:X\to M_n$ into the algebra $M_n$ of all $n\times n$ complex matrices?
I'm assuming your compact space $X$ is Hausdorff. Then the answer is yes.
Suppose $I$ is a closed two-sided ideal in $C(X, M_n)$.
For each $x \in X$, $I(x) = \{f(x): f \in I\}$ is a two-sided ideal in $M_n$. But
$M_n$ is a simple ring: it has no two-sided ideals except itself and $\{0\}$.
Let $S = \{x \in X: \; I(x) = \{0\}\}$. Then $I \subseteq I(S) = \{f \in C(X, M_n):\; f|_S = 0\}$.
Now suppose $f \in I(S)$. I need to show $f \in I$. For each $x \notin S$, $I(x) = M_n$ so there is some $g_x \in I$ with $g_x(x) = f(x)$. Given $\epsilon > 0$, we can find a finite partition of unity $\rho_1, \ldots, \rho_N$ of $X$ and $g_1, \ldots, g_N \in I$ such that $\|f(x) - \sum_j \rho_j(x) g_j(x)\| < \epsilon$ for all $x \in X$. Since $I$ is closed, $f \in I$.
