How to calculate the $K_0$ and $K_1$ groups for $A$ 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. 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?
 A: One can describe the algebra $A$ as the unitization of $M_n(B)$ where $B = C_0[0,1)$. Hence,
$$
K_0(A) \cong K_0(M_n(B))\oplus K_0(\mathbb{C})
$$
But $K_0(M_n(B)) \cong K_0(B)$ by the stability of $K_0$, and $K_0(B) = 0$ since the identity map on $B$ is homotopic to the zero map. Thus, 
$$
K_0(A)\cong \mathbb{Z}
$$
A: $\require{AMScd}$
If you want to use six-term exact sequence, one can consider the followings short exact sequence:
$$
0 \to CM_n \overset \imath \to A \overset {\mathrm {ev}} \to M_n \to 0,
$$
where $CM_n$ is the cone over $M_n$, $\imath$ the inclusion and $\mathrm{ev}$ the  evaluation map at $0$. 
\begin{CD}
K_0(CM_n) @>> > K_0(A) @>> > K_0(M_n)\\
@AAA @. @VVV \\
K_1(M_n) @<< < K_1(A) @<< < K_1(CM_n)
\end{CD}
Since the cone over any algebra is contractible, and $K_1(M_n)$ is trivial, this becomes 
\begin{CD}
0 @>> > K_0(A) @>> > K_0(M_n)\\
@AAA @. @VVV \\
0  @<< < K_1(A) @<< < 0
\end{CD}
So we have an isomorphism
$$
K_0(\mathrm{ev}) \colon K_0(A) \to K_0(M_n) \cong \mathbb Z
$$
and $K_1(A) = \{0\}$.
Remark: 
The same proof works if you look at functions $f \colon [0,1] \to M_n$ such that $f(0)$ is diagonal.
