Suppose we wish to write the vector
\[\begin{pmatrix}1\\4\\-3\end{pmatrix}\]
in terms of the vectors \[\begin{pmatrix}1\\-2\\5\end{pmatrix} , \: \begin{pmatrix}2\\-3\\0\end{pmatrix} , \: \begin{pmatrix}0\\1\\3\end{pmatrix}\]
Then
\[\begin{pmatrix}1\\4\\-3\end{pmatrix} = a \begin{pmatrix}1\\-2\\5\end{pmatrix} +b \begin{pmatrix}2\\-3\\0\end{pmatrix} +c \begin{pmatrix}0\\1\\3\end{pmatrix}\]
This is equivalent to the system of equations
\[1=a+2b\]
\[4=-2a-3b+c\]
\[-3=5a+c\]
This system has solutions
\[a=- \frac{17}{11} ,\: b=\frac{14}{11} , \: c=\frac{52}{11} \]
Then
\[\begin{pmatrix}1\\4\\-3\end{pmatrix} = - \frac{17}{11} \begin{pmatrix}1\\-2\\5\end{pmatrix} + \frac{14}{11} \begin{pmatrix}2\\-3\\0\end{pmatrix} + \frac{52}{11} \begin{pmatrix}0\\1\\3\end{pmatrix}\]
We can write define a basis
\[\{ \mathbf{e_1}, \: \mathbf{e_2}, \: \mathbf{e_3} \}\]
then\[\mathbf{e_1} = \begin{pmatrix}1\\-2\\5\end{pmatrix} , \: \mathbf{e_2}= \begin{pmatrix}2\\-3\\0\end{pmatrix} , \: \mathbf{e_3} =\begin{pmatrix}0\\1\\3\end{pmatrix}\]
Then
\[\begin{pmatrix}1\\4\\-3\end{pmatrix} = - \frac{17}{11} \mathbf{e_1} + \frac{14}{11} \mathbf{e_2} + \frac{52}{11} \mathbf{e_3}\]