Solution: Let \[ v_1 = \begin{bmatrix} 1 \\ 2 \\ \end{bmatrix}, \text{ and } v_2 = \begin{bmatrix} 3 \\ 5 \\ \end{bmatrix}. \] We need to find $\alpha ,\beta \in \mathbb{R}$ such that \[ \alpha \begin{bmatrix} 1 \\ 2 \\ \end{bmatrix}+ \beta \begin{bmatrix} 3 \\ 5 \\ \end{bmatrix} = \begin{bmatrix} -2 \\ 4 \\ \end{bmatrix}. \] Equivalently, we want to solve the system of linear equations \begin{align*} \alpha + 3\beta & = -2 \\ 2\alpha + 5\beta & = 4. \end{align*} Solving the above two equations, give $\alpha = 22$ and $\beta =-8$. Therefore, the coordinates in terms of the basis $B$ are \[ \begin{bmatrix} 22 \\ -8 \\ \end{bmatrix}. \]

At first, we will find the image of the basis vectors under the linear transformation $T$. \begin{align*} w_1 = T \begin{bmatrix} 1 \\ 2 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ -1 \\ \end{bmatrix}, \text{ and } w_2 = T \begin{bmatrix} 3 \\ 5 \\ \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}. \end{align*} One way to find the matrix of the linear transformation is to write the vectors $w_1$ and $w_2$ in terms of $v_1$ and $v_2$. The matrix of $T$ will be given by the coordinates for the vectors $w_1$ and $w_2$. Another way is to apply the change of the basis matrix. It is easy to find the matrix with respect to the standard basis, say $T_s$, and then the matrix with respect to $B$ will be \[ T_B = \begin{bmatrix} T_s ^{-1} \left( w_1 \right) & T_s ^{-1} \left( w_2 \right) \\ \end{bmatrix}. \] Here we have \[ T_s ^{-1} = \begin{bmatrix} 1 & 3 \\ 2 & 5 \\ \end{bmatrix}^{-1} = \begin{bmatrix} -5 & 3 \\ 2 & -1 \\ \end{bmatrix}. \] Therefore, \begin{align*} T_B & = \begin{bmatrix} T_s ^{-1} \left( w_1 \right) & T_s ^{-1} \left( w_2 \right) \\ \end{bmatrix} \\ & = \begin{bmatrix} -3 & -8 \\ 1 & 3 \\ \end{bmatrix}. \end{align*}