1
2
3
4
Correct answer is C
\(\begin{vmatrix} k & k \\ 4 & k \end{vmatrix} + \begin{vmatrix} 2 & 3 \\ -1 & k \end{vmatrix} = 6\)
\(\begin{vmatrix} k & k \\ 4 & k \end{vmatrix} = (k^{2} - 4k)\)
\(\begin{vmatrix} 2 & 3 \\ -1 & k \end{vmatrix} = (2k + 3)\)
\(\therefore (k^{2} - 4k) + (2k + 3) = k^{2} - 2k + 3 = 6\)
\(k^{2} - 2k - 3 = 0\), factorising, we have \(k + 1 = 0\) or \(k - 3 = 0\)
Since k > 0, k = 3.