Sure! Let's work through the problem step-by-step to find \(4a - 3b + c\), given the vectors \(a\), \(b\), and \(c\).
The vectors are given as:
[tex]\[
a = \begin{pmatrix} 1 \\ 3 \\ -1 \end{pmatrix}, \quad b = \begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix}, \quad c = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}
\][/tex]
First, let's calculate \(4a\):
[tex]\[
4a = 4 \cdot \begin{pmatrix} 1 \\ 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 4 \cdot 1 \\ 4 \cdot 3 \\ 4 \cdot -1 \end{pmatrix} = \begin{pmatrix} 4 \\ 12 \\ -4 \end{pmatrix}
\][/tex]
Next, calculate \(3b\):
[tex]\[
3b = 3 \cdot \begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix} = \begin{pmatrix} 3 \cdot 2 \\ 3 \cdot 3 \\ 3 \cdot -4 \end{pmatrix} = \begin{pmatrix} 6 \\ 9 \\ -12 \end{pmatrix}
\][/tex]
Now, let's subtract \(3b\) from \(4a\):
[tex]\[
4a - 3b = \begin{pmatrix} 4 \\ 12 \\ -4 \end{pmatrix} - \begin{pmatrix} 6 \\ 9 \\ -12 \end{pmatrix} = \begin{pmatrix} 4 - 6 \\ 12 - 9 \\ -4 - (-12) \end{pmatrix} = \begin{pmatrix} -2 \\ 3 \\ 8 \end{pmatrix}
\][/tex]
Finally, add \(c\) to the result:
[tex]\[
4a - 3b + c = \begin{pmatrix} -2 \\ 3 \\ 8 \end{pmatrix} + \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 + 1 \\ 3 - 1 \\ 8 + 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \\ 9 \end{pmatrix}
\][/tex]
Thus, the result of \(4a - 3b + c\) is:
[tex]\[
\boxed{\begin{pmatrix} -1 \\ 2 \\ 9 \end{pmatrix}}
\][/tex]
