Answer
12 square units
Work Step by Step
$\textbf{S}$ is a parallelogram determined by two vectors;
Let the vectors be $\textbf{vector1}\,and\, \textbf{vector2}$.
$\textbf{vector1} =\begin{bmatrix}4\\-7\end{bmatrix}$
$\textbf{vector2} =\begin{bmatrix}0\\1\end{bmatrix}$
area of parallelogram, $=|det[{\textbf{vector 1} \textbf{vector 2}}]$|
$area\,of\,\textbf{S}=|det\begin{bmatrix}4&0\\-7&1\end{bmatrix}|=|4\times1-(-7\times0)|=4$
Given T as a linear transformation and S as a set in the T domain and
$\textbf{A}=\begin{bmatrix}5&2\\1&1\end{bmatrix}$
To compute the area of the image of S;
Let T(S) be a set of images of points in S
$area\, of \,T(S) = |det A|.{\{area\, of\, S\}}=|det
\begin{bmatrix}5&2\\1&1\end{bmatrix}|*4$
$area\, of \,T(S) =(|5\times1-2\times1|)*4=3*4=12$ square units