Step 1 :First, we try to express \(\begin{pmatrix} 5 \\ -1 \\ 2 \\ -4 \end{pmatrix}\) as a linear combination of \(\begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} -1 \\ 1 \\ 0 \\ 2 \end{pmatrix}\):
Step 2 :\(\begin{pmatrix} 5 \\ -1 \\ 2 \\ -4 \end{pmatrix} = a \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} + b \begin{pmatrix} -1 \\ 1 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} a - b \\ a + b \\ a \\ a + 2b \end{pmatrix}\)
Step 3 :Solving the system of equations \(a - b = 5\), \(a + b = -1\), \(a = 2\), and \(a + 2b = -4\), we find that \(a = 2\) and \(b = -3\).
Step 4 :Thus, \(\begin{pmatrix} 5 \\ -1 \\ 2 \\ -4 \end{pmatrix} = 2 \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} - 3 \begin{pmatrix} -1 \\ 1 \\ 0 \\ 2 \end{pmatrix}\).
Step 5 :By the linearity of the transformation \(T\), we have \(T \begin{pmatrix} 5 \\ -1 \\ 2 \\ -4 \end{pmatrix} = 2 T \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} - 3 T \begin{pmatrix} -1 \\ 1 \\ 0 \\ 2 \end{pmatrix}\).
Step 6 :Substituting the given values of \(T\begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}\) and \(T\begin{pmatrix} -1 \\ 1 \\ 0 \\ 2 \end{pmatrix}\), we get \(T \begin{pmatrix} 5 \\ -1 \\ 2 \\ -4 \end{pmatrix} = 2 \begin{pmatrix} 5 \\ 1 \\ -3 \end{pmatrix} - 3 \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}\).
Step 7 :Calculating the result, we find \(T \begin{pmatrix} 5 \\ -1 \\ 2 \\ -4 \end{pmatrix} = \boxed{\begin{pmatrix} 4 \\ 2 \\ -7 \end{pmatrix}}\).