use **similarity transformation** to find the matrix $ \ B \ $ with respect to the basis $ \ \{(1,2) , \ (3,5) \} \ \ of \ \ \mathbb{R}^2 $ . Consider the linear operator $ \ T(x,y)=(5x,x-y) \ \ on \ \ \mathbb{R}^2 $ .  Find the matrix $ \ A \ $ of  $ \ T \ $ relative to the standard basis on $ \ \mathbb{R}^2 \ $ . Then use similarity transformation to find the matrix $ \ B \ $ with respect to the basis $ \ \{(1,2) , \ (3,5) \} \ \ of \ \ \mathbb{R}^2 $ . 
Note: The answer is given as $ B =\begin{pmatrix}-29 & -56 \\ 17 & 33 \end{pmatrix} $
Answer:
I have found $ \ A= \begin{pmatrix} 5 & 0 \\ 1 & -1 \end{pmatrix} $ . 
Now $ \ B=P^{-1} AP \ $  , 
where $ \ P \ $ is the transition matrix from the basis $ \ \{(1,2) , (3,5) \} \ $  to the basis $ \ \{(1,0) , (0,1) \} \ $. 
Now calculating I got , 
$ \ P=\begin{pmatrix}1 & 3 \\ 2 & 5 \end{pmatrix} $
Thus, 
$ \ P^{-1} =\begin{pmatrix}-5 & 2 \\ 3 & -1 \end{pmatrix} $ 
Hence , 
$ B=P^{-1} AP = \begin{pmatrix}-5 & 2 \\ 3 & -1 \end{pmatrix}  \begin{pmatrix} 5 & 0 \\ 1 & -1 \end{pmatrix}  \begin{pmatrix} 1 & 3 \\ 2 & 5 \end{pmatrix} = \begin{pmatrix}-27 & -79 \\ 16 & 47 \end{pmatrix} $
 $
But the answer does not match with the given answer .
Was my calculation and appoach right ?
If not then how to solve the problem ?
Is there any help ?

You made a "silly" minor  mistake $$P=\left(
\begin{array}{ll}
 1 & 2 \\
 3 & 5 \\
\end{array}
\right)$$
If you use that then $$ P^{-1}AP=\left(
\begin{array}{cc}
 -5 & 2 \\
 3 & -1 \\
\end{array}
\right).\left(
\begin{array}{cc}
 5 & 0 \\
 1 & -1 \\
\end{array}
\right).\left(
\begin{array}{cc}
 1 & 2 \\
 3 & 5 \\
\end{array}
\right)=\left(
\begin{array}{cc}
 -29 & -56 \\
 17 & 33 \\
\end{array}
\right)$$
