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(D) Given $A = \begin{bmatrix} 1 & 3 \ 4 & -3 \end{bmatrix}$ and $S = \left\{ \begin{bmatrix} x \ y \end{bmatrix} \in \mathbb{R}^2 \mid A \begin{bma

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Uncountable

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(D) Given $A = \begin{bmatrix} 1 & 3 \ 4 & -3 \end{bmatrix}$ and $S = \left\{ \begin{bmatrix} x \ y \end{bmatrix} \in \mathbb{R}^2 \mid A \begin{bmatrix} x \ y \end{bmatrix} = 3 \begin{bmatrix} x \ y \end{bmatrix} ight\}$.To find the cardinality of $S$,we solve the equation $A \begin{bmatrix} x \ y \end{bmatrix} = 3 \begin{bmatrix} x \ y \end{bmatrix}$.$\begin{bmatrix} 1 & 3 \ 4 & -3 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 3x \ 3y \end{bmatrix}$$\begin{bmatrix} x + 3y \ 4x - 3y \end{bmatrix} = \begin{bmatrix} 3x \ 3y \end{bmatrix}$This gives the system of equations:$x + 3y = 3x \implies 2x = 3y \implies x = \frac{3}{2}y$$4x - 3y = 3y \implies 4x = 6y \implies 2x = 3y \implies x = \frac{3}{2}y$Both equations reduce to the same condition $x = \frac{3}{2}y$.Thus,$S = \left\{ \begin{bmatrix} \frac{3}{2}y \ y \end{bmatrix} \mid y \in \mathbb{R} ight\} = \left\{ y \begin{bmatrix} 1.5 \ 1 \end{bmatrix} \mid y \in \mathbb{R} ight\}$.Since $y$ can be any real number,there are infinitely many such vectors.The set $S$ represents a line in $\mathbb{R}^2$ passing through the origin,which contains uncountably many points.Therefore,the cardinality of $S$ is uncountable.

Let $A = \begin{bmatrix} 1 & 3 \\ 4 & -3 \end{bmatrix}$. Let $S = \left\{ \begin{bmatrix} x \\ y \end{bmatrix} \in \mathbb{R}^2 \mid A \begin{bmatrix} x \\ y \end{bmatrix} = 3 \begin{bmatrix} x \\ y \end{bmatrix} \right\}$. What is the cardinality of $S$?

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