Find a value of k so that the matrix \[ A= \left[ \begin{array}{cccc} 2 & 4 \\ 3 & k \\ \end{array} \right] \\ \] has det(A) = 0.
det(A) = (2)(k) - (4)(3)
so det(A) = 0 implies 2k - 12 = 0
which mean 2k = 12
hence k=6.
So the det(A) = 0 if and only if k=6