linear transformation matrix Any help on this linear transformation question is very much appreciated.
Let $V$ denote the real vector space $R^2$ and $\psi  : V \rightarrow V$ be a real linear transformation such
that $\psi ((1, 0)) = (11, 8)$ and $\psi ((0, 1)) = (4, 3)$. Express the image $\psi ((x, y))$ of $(x, y)$ in terms
of $x$ and $y$.
Assume that $w_1 = (4, 5)$ and $w_2 = (9,11)$ form an ordered basis $B$ for $V$ . Working from the
denition determine the matrix $M^B_B (\psi)$   with respect to the basis $B$.
does $ M^B_B(\psi)=
\begin{bmatrix}
-469 & -1048 \\
388 & 867
\end{bmatrix}$ ?
thanks in advance for any help
 A: The answer to the first part of the question is implicit in Paul's answer, but I'll bring it out explicitly. 
$(x,y)=x(1,0)+y(0,1)$, and $\psi$ is linear, so $$\psi(x,y)=x\psi(1,0)+y\psi(0,1)=x(11,8)+y(4,3)=(11x+4y,8x+3y)$$
A: Denote $E$ the standard basis $(1,0)$, $(0,1)$. By definition, the matrix $M^E_E(\psi)$ (using your notation) is given by 
$$M^E_E(\psi)=
\begin{bmatrix}
11 & 4 \\
8 & 3
\end{bmatrix}.$$
Now, the transition matrix from $B$ to $E$ is given by
$$M^E_B(id)=
\begin{bmatrix}
4 & 9 \\
5 & 11
\end{bmatrix}.$$
Hence the transition matrix from $E$ to $B$ is given by
$$M^B_E(id)=
\begin{bmatrix}
4 & 9 \\
5 & 11
\end{bmatrix}^{-1}=\frac{1}{-1}\begin{bmatrix}
11 & -9 \\
-5 & 4
\end{bmatrix}=\begin{bmatrix}
-11 & 9 \\
5 & -4
\end{bmatrix}.$$
Combining all these, the matrix $M^B_B(\psi)$ is given by 
$$M^B_B(\psi)=M^B_E(id)\cdot M^E_E(\psi)\cdot M^E_B(id)=\begin{bmatrix}
-11 & 9 \\
5 & -4
\end{bmatrix}\cdot\begin{bmatrix}
11 & 4 \\
8 & 3
\end{bmatrix}\cdot\begin{bmatrix}
4 & 9 \\
5 & 11
\end{bmatrix}=...$$
