Given
$$
\begin{aligned}
& A=\left[\begin{array}{cc}
1 & 3 \\
4 & -3
\end{array}\right] \\
& S=\left\{\left[\begin{array}{l}
x \\
y
\end{array}\right] \in R^2 \mid A\left[\begin{array}{l}
x \\
y
\end{array}\right]=3\left[\begin{array}{l}
x \\
y
\end{array}\right]\right\}
\end{aligned}
$$
To Find Cardinality of $S$
$$
\begin{array}{lr}
\because & A\left[\begin{array}{l}
x \\
y
\end{array}\right]=3\left[\begin{array}{l}
x \\
y
\end{array}\right] \\
& {\left[\begin{array}{cc}
1 & 3 \\
4 & -3
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
3 x \\
3 y
\end{array}\right]} \\
\Rightarrow & {\left[\begin{array}{c}
x+3 y \\
4 x-3 y
\end{array}\right]=\left[\begin{array}{l}
3 x \\
3 y
\end{array}\right]}
\end{array}
$$
$$
\begin{aligned}
& \Rightarrow x+3 y=3 x \text { and } 4 x-3 y=3 y \\
& \Rightarrow \quad 2 x=3 y \\
& x=3 y / 2 \\
& \therefore \quad S=\left\{\left[\begin{array}{l}
\frac{3 y}{2} \\
y
\end{array}\right] \in R^2 \mid A\left[\begin{array}{l}
x \\
y
\end{array}\right]=3\left[\begin{array}{l}
x \\
y
\end{array}\right]\right\} \\
& =\left\{\left[\begin{array}{c}
\frac{3}{2} \\
1
\end{array}\right] \in R^2 \mid A\left[\begin{array}{l}
x \\
y
\end{array}\right]=3\left[\begin{array}{l}
x \\
y
\end{array}\right]\right\} \\
& \because \quad y \in R \\
& \therefore|S|=R \\
&
\end{aligned}
$$
$\Rightarrow$ Cardinality of $S$ is uncountable.