Consider the group homomorphism $\phi:M_{2}(\mathbb{R})\to \mathbb{R}$ given by $\phi(A)=Tr(A)$. Then the kernel of $\phi$ is not isomorphic to
(a)$M_{2}\mathbb{R}/\{A\in M_{2}(\mathbb{R}):\phi(A)=0\}$
(b)$\mathbb{R^{2}}$
(c)$\mathbb{R^{3}}$
(d)$Gl_{2}(\mathbb{R})$
Here the correct answer is given as option (a). I am not getting this how they got that. And if option (a) is correct then the kernel is isomorphic to $\mathbb{R^{3}}$, $\mathbb{R^{2}}$ and $Gl_{2}(\mathbb{R})$. But how the kernel will be isomorphic to this three sets.
Since the kernel is the set $\{\begin{bmatrix}a&&b\\c&&-a\end{bmatrix}\in M_{2}(\mathbb{R}):a,b,c \in \mathbb{R}\}$, for option $c$, I can define the isomorphism map as $f\Big(\begin{bmatrix}a&&b\\c&&-a\end{bmatrix}\Big)=(a,b,c)$. But what should be the map for the other two options, I am not getting and why the option a is correct, that also I want to know.
Thanks in advance...