Chap D, M10, p:271

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.

Solution

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