Answer:
We can write the given system of equations as a matrix equation:
$\begin{bmatrix} 2 & 4 \ 4 & k \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}$
To find nontrivial solutions (i.e., $x$ and $y$ not both equal to zero), the coefficient matrix must be singular, which means its determinant must be zero:
$\det\begin{bmatrix} 2 & 4 \ 4 & k \end{bmatrix} = 2k - 16 = 2(k - 8) = 0$
Thus, $k = 8$ is the only value for which there exists a nonzero 2-dimensional vector $\boldsymbol{v} = \begin{bmatrix} x \ y \end{bmatrix}$ satisfying the given system of equations. For $k \\eq 8$, the only solution is the trivial one, $\boldsymbol{v} = \boldsymbol{0}$.
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