Consider the tuple
\(\begin{bmatrix}
s + 2t \\
s - 2t \\
3s + 4t
\end{bmatrix}\).
It can be written as
\(s~\begin{bmatrix} a\\1\\b \end{bmatrix} +
t~\begin{bmatrix} c\\d \\ 4\end{bmatrix}\)
What is \(a+b+c+d\)?
The answer is 4.
The tuple can be written as
\(s~\begin{bmatrix} 1\\1\\3 \end{bmatrix} +
t~\begin{bmatrix} 2\\-2 \\ 4\end{bmatrix}.\)
So \(a = 1\), \(b = 3\), \(c = 2\), and \(d = -2\), giving
\(a + b+ c +d = 1+3 +2 +(-2) = 4\).