(2s+3)([x y]) not sure exactly what we are supposed to do with the 3

[karush]

ok I posted this on another thread
$(2S+3T)\left(\left[\begin{array}{c}x \\ y \end{array}\right]\right)
= \left[\begin{array}{c}
2-4y \\ 6x-2y
\end{array}\right]
+3\left[\begin{array}{c}x+y \\ x-y\\2x+3y \end{array}\right]
=\left[\begin{array}{c}
2-4y \\ 6x-2y
\end{array}\right]
+\left[\begin{array}{c}3x+3y \\ 3x-3y\\6x+9y \end{array}\right]$ but the teacher said it was supposed to be $(2S+{\color{red}{\textbf{3}}})\left(\left[\begin{array}{c}x \\ y \end{array}\right]\right)$
instead ... not sure exactly what we are supposed to do with the ${\color{red}{\textbf{3}}}$
2S was formerly calculated just c/p here

 

[topsquark]

ok I posted this on another thread
$(2S+3T)\left(\left[\begin{array}{c}x \\ y \end{array}\right]\right)
= \left[\begin{array}{c}
2-4y \\ 6x-2y
\end{array}\right]
+3\left[\begin{array}{c}x+y \\ x-y\\2x+3y \end{array}\right]
=\left[\begin{array}{c}
2-4y \\ 6x-2y
\end{array}\right]
+\left[\begin{array}{c}3x+3y \\ 3x-3y\\6x+9y \end{array}\right]$ but the teacher said it was supposed to be $(2S+{\color{red}{\textbf{3}}})\left(\left[\begin{array}{c}x \\ y \end{array}\right]\right)$
instead ... not sure exactly what we are supposed to do with the ${\color{red}{\textbf{3}}}$
2S was formerly calculated just c/p here

You are overthinking again, I think.
\(\displaystyle 3 \left [ \begin{matrix} x \\ y \end{matrix} \right ] = \left [ \begin{matrix} 3x \\ 3y \end{matrix} \right ] \)

-Dan

 

[karush]

sorry had a "duh" moment

 

[HOI]

A better way to write "2S+ 3" is "2S+ 3I" where "I" is the identity operator, Iv= v.