Find coordinate vector for change of bases when matrix is non-square Find $x_{b2}$ if $x_{b1} = \begin{pmatrix}2 \\ 5  \end{pmatrix}_{b1}$
Let $$ B_{1} = \begin{pmatrix}
3 & -1 \\
-5 & 4 \\
5 & -3
\end{pmatrix}\, 
\ \;\;\;\;\;\;\;\;\;\; B_{2}=
\begin{pmatrix} 
2 & 1 \\
-1 & 3 \\
2 & -1
\end {pmatrix}
$$
I know I have to use the equation $x_{b2} = V^{-1}Ux_{b1}$, where  $V =B_{2}$ by V is not invertible since it is not a square matrix. The textbook provides a solution in the back of the book, but no examples on how to deal with non-square matrices. How do I solve these type of questions.

The coordinates of your vector, $x$, relative to the basis $B_1$ are $\begin{bmatrix} 2 \\ 5 \end{bmatrix}$. This means that 
$$ x = B_1 \begin{bmatrix} 2 \\ 5 \end{bmatrix} = 2\begin{bmatrix} 3 \\ -5 \\ 5 \end{bmatrix} + 5\begin{bmatrix} -1 \\ 4 \\ -3 \end{bmatrix} = \begin{bmatrix} 1 \\ 10 \\ -5 \end{bmatrix}$$
To find $x$'s coordinates relative to the other basis, you need to solve the system:
$$ B_2 \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ -1 & 3 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 1 \\ 10 \\ -5 \end{bmatrix}$$
