If (begin{vmatrix} k & k \ 4 & k end{vmatrix} + begin{vmatrix} 2 & 3 \ -1 & k end{vmatrix} = 6), find the value of the constant k, where k > 0.
-
A.
1 -
B.
2 -
C.
3 -
D.
4
Correct Answer: Option C
Explanation
(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.