How to determine which of the following matrices are similar? If we have the following three matrices: 
$$
A=\begin{bmatrix}
7 &1 \\ 
-5 &3 
\end{bmatrix},\;\;
B=\begin{bmatrix}
5 &-1 \\ 
1 &5 
\end{bmatrix},\;\;
C=\begin{bmatrix}
5 &1 \\ 
1 &5 
\end{bmatrix}.
$$
What is the right procedure to determine if matrices are similar? I know that if the matrices are similar, then the matrices have the samen eigenvalues. Using the negative of this (determine the eigenvalues of each matrix, if they are not similar they cannot be similar), I got that only $A$ and $B$ potentially could be similar. 
I also know that if the 2 matrices $A$ and $B$ are similar then $A=PBP^{−1}$ but I don't know how to work this out further. 
So how do I show that matrices are similar, using $A=PBP^{-1}$ for example to show it for $A$ and $B$?
 A: As similar matrices have similar determinants, matrix $C$ is not similar to $A$ or $B$ as $\det C = 24$ but $\det A = \det B = 26$. 
Now we notice that $A$ and $B$ have the similar egeinvalues from characteristic equation 
$$
\lambda^2 - 10\lambda + 26 = 0
$$
which gives
$$
\lambda_{1,2} = 5 \pm i.
$$
We may diagonilize bouth matrises $A$ and $B$ and get that $A = PDP^{-1}$ and $B = QDQ^{-1}$ where 
$$
D = \begin{pmatrix}
5 - i & 0\\
0 & 5 + i
\end{pmatrix}.
$$
Now one may see that matrix $T = QP^{-1}$ satisfies equation $B = TAT^{-1}$:
$$
T = 
\begin{pmatrix}
-i & i \\ 1 & 1
\end{pmatrix}
\begin{pmatrix}
-\frac{5i}{2} & \frac{1}{2}-i \\ \frac{5i}{2} & \frac{1}{2}+i
\end{pmatrix} =
\begin{pmatrix}
-5 & -2 \\ 0 & 1
\end{pmatrix},
$$
$$
T^{-1} = 
\frac{1}{5}\begin{pmatrix}
-1 & -2 \\ 0 & 5
\end{pmatrix},
$$
$$
TAT^{-1} = 
\frac{1}{5}\begin{pmatrix}
-5 & -2 \\ 0 & 1
\end{pmatrix}\begin{pmatrix}
7 & 1 \\ -5 & 3
\end{pmatrix}\begin{pmatrix}
-1 & -2 \\ 0 & 5
\end{pmatrix} = \begin{pmatrix}
5 & -1 \\ 1 & 5
\end{pmatrix} = B.
$$
A: $A$ and $B$ cannot be similar to $C$, since $\det(A)=\det(B)=26\neq \det(C)=24$.
On the other hand $A$ and $B$ are similar with 
$$
P=\begin{pmatrix} 2 & 2 \cr 6 & 2 \end{pmatrix}.
$$
Here we have $B=PAP^{-1}$. I do not need eigenvalues for this computation. I start with an invertible matrix $P=\begin{pmatrix} s_1 & s_3 \cr s_2 & s_4 \end{pmatrix}$ and then rewrite $BP=PA$ as 
$$
2s_1 + s_2 - 5s_3=s_1 - 2s_3 + s_4= - s_1 + 2s_2 - 5s_4=s_2 - s_3 - 2s_4=0.
$$
The determinant condition is $s_1s_4-s_2s_3\neq 0$. Then we can solve the linear equations and take one with nonzero determinant.
