Conjugate elements of $GL_2(\mathbb{R})$ Decide whether or not the two matrices $A= \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix}$ and $B= \begin{pmatrix} 1 & 1 \\ -2 & 4 \end{pmatrix}$ are conjugate elements of the general linear group $GL_2(\mathbb{R})$
So I need to show that there exists a matrix $X= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in GL_2(\mathbb{R})$ such that $a,b,c,d \in \mathbb{R}$ and such that $ad-bc \neq 0$  
I solved the system $XAX^{-1}=B$ and found that the the matrix is of the form $X= \begin{pmatrix} a & 0 \\ b & 0 \end{pmatrix}$ which is not invertible hence A and B are not conjugate elements.
Is it correct?  Is there an easier way to solve this type of question ?
 A: It can readily be seen that $2$ is an eigenvalue of $B$ because the sum of each row is always $2$. Since the trace of $B$ is $5$ it follows that $3$ is also an eigenvalue of $B$, therefore $B$ is diagonalizable and similar to $A$, that is, there exists $X$ such that $X^{-1}BX=A$.
So you have made a miscalculation in solving the system. If you want to solve a system I suggest solving the equivalent, and in my opinion easier system:$XA=BX$.
Edit: Diagonalizability is a field-dependent concept. All that was said above is with respect to $\Bbb R$. It maybe the case that a real matrix isn't diagonalizable in $\Bbb R$ and it is $\Bbb C$.
A: $XA=BX$ is equivalent to $c=2a$ and $d=b$, i.e.,
$$
X=\begin{pmatrix} a & b \\
2a & b
\end{pmatrix}
$$
We need $0\neq \det(X)=-ab$, which is possible with, say, $a=b=1$. 
Remark: It is easy to see that $A$ and $B$ have the same characteristic polynomial, since they have the same trace and the same determinant. For $GL_2(\mathbb{Z})$ this is not enough to conclude that they are similar.
A: Because the trace and the determinant coincide, so does the characteristic polynomial, in particular its roots=eigenvalues. Both eigenvalues are distinct, so both matrices similar to a diagonal matrices with entries $2$ and $3$. 
