Let $A=\{f\in C([0,1],M\_n)\mid f(0)$ is scalar matrix $\}$.

Then find the $K\_0(A)$ and $K\_1(A)$.

I am trying to use the SES $J \rightarrow A \rightarrow A/J$ where $J$ can be taken as some closed ideals of $A$. Such a closed ideals may be found in a similar way due to my previously asked question [here](https://math.stackexchange.com/questions/2883178/show-that-the-space-of-continuous-functions-on-the-compact-hausdorff-space-with). Now $J$ contains all the continuous functions that are zero on some closed subsets of $[0,1]$, that is, finite points or the closed subintervals of $[0,1]$. Then it seems to me that $M\_n(J)$ consists of those matrices with operator-entries that is actually zero in above case. These matrices have zero diagonals and therefore $K\_0(J)=0$. But after that I don't know how to do apart from $K\_0(A)\simeq K\_(A/J)$ by the half-exactness of SES. 

I am also assuming that there should be a way to consider the $K$-groups by definition, for example, when calculating $K(\mathbb {C})$, take the difference of $[p]\_0-[q]\_0$, each of which is the equivalence class of matrices in $M\_n(\mathbb {C})$. Then the generators are just those matrices with different numbers of $1$ in the diagonal, plus all zeros one; so $V(A)$= $\mathbb{N} \cup {0}$. Counting the difference, we have $K\_0(\mathbb{C})=\mathbb {Z}$.

Could anybody show me the technique for calculating $K\_1$ and $K\_0$? Are there any standard techniques to calculate $K$-groups in general?