Answer
12 square units
Work Step by Step
S is a parallelogram determined by two vectors;
$\textbf{vector 1}=\begin{bmatrix}-2\\3\end{bmatrix}$
$\textbf{vector 2}=\begin{bmatrix}-2\\5\end{bmatrix}$
area of parallelogram, $=|det[{\textbf{vector 1} \textbf{vector 2}}]$|
$area=|det\begin{bmatrix}-2&-2\\3&5\end{bmatrix}|=|-10+6|=4$
Given T as a linear transformation and S as a set in T domain
and $\textbf{A}=\begin{bmatrix}6&-3\\-3&2\end{bmatrix}$
To compute the area of the image of S;
Let T(S) be a set of images of points in S
$area\, of \,T(S) = |det A|.{\{area\, of\, S\}}=|det \begin{bmatrix}6&-3\\-3&2\end{bmatrix}|*4$
$area\, of \,T(S) =(|12-9|)*4=3*4=12$ square units
