$C^*$-subalgebra of $M_2(\mathbb{C})$ Consider the following subalgebra of $M_2(\mathbb{C})$:
$$A=
\left\{\begin{bmatrix}
   a & b \\
   b & a
  \end{bmatrix} : a,b\in \mathbb{C}\right\}.$$
One method I know that $A$ is isometrically *-isomorphic to $C(K)$, where $K=\{1,2\}$. (Because we can assume function $f$ such that $f(1)= \begin{bmatrix}
   1 & 0 \\
   0 & 1
  \end{bmatrix}$
 and 
$f(2)=\begin{bmatrix}
   0 & 1 \\
   1 & 0
  \end{bmatrix}$, then we use the Gelfand transform between $C(K)$ and $C(\mathscr{M})$, where $\mathscr{M}$ is the maximal ideals of  $C(K)$. Hence, using the Gelfand transform $\widehat{f}(J)=f(x), J\in \mathscr{M}, x\in K $ to get the maximal ideals of $A$.
But this problem is asked to show that $\Delta=\{\phi_{1}, \phi_{2}\}$, where $\phi_{1}(y)=a+b, \phi_{2}(y)=a-b$. Here, $\Delta$ is the set of all complex homomorphism  $A \to \mathbb{C}$.
$\textbf{How to solve this problem in this way? }$

Suppose that $\pi:A\to\mathbb C$ is a homorphism. 
If $\pi(I)=0$, then $\pi=0$. Otherwise, from $\pi(I)=\pi(I^2)=\pi(I)^2$, we deduce that $\pi(I)=1$. In any case, $\pi(aI)=a$.  Also, since 
$$
x=\begin{bmatrix} 0&1\\1&0\end{bmatrix}
$$
satisfies $x^2=I$, we deduce that $\pi(x)^2=1$. So $\pi(x)=1$ or $\pi(x)=-1$. And those are all the choice we have: when $\pi(x)=1$, we obtain the homomorphism 
$$
\pi\left(\begin{bmatrix} a&b\\b&a\end{bmatrix} \right)=\pi(aI+bx)=a+b. 
$$
When $\pi(x)=-1$, we obtain 
$$
\pi\left(\begin{bmatrix} a&b\\b&a\end{bmatrix} \right)=\pi(aI+bx)=a-b. 
$$
