Given \(\begin{vmatrix} 2 & -3 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} -6 \\ k \end{vmatrix}  \begin{vmatrix} 3 \\ -26 \end{vmatrix} = 15\). Solve for k.

A.

-8

B.

-5

C.

-4

D.

-3

Correct answer is B

\(\begin{vmatrix} 2 & -3 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} -6 \\ k \end{vmatrix}  \begin{vmatrix} 3 \\ -26 \end{vmatrix} = 15\)

 

\(\begin{vmatrix} 2[-6] & - 3k \\ 1[-6] & + 4k \end{vmatrix} = \begin{vmatrix} 3 \\ -26 \end{vmatrix}\)

 

\(\begin{vmatrix} -12 & - 3k \\ -6 & + 4k \end{vmatrix} = \begin{vmatrix} 3 \\ -26 \end{vmatrix}\)

 

-12 - 3k = 3

-3k = 3 + 12

k = \(\frac{15}{-3}\)

k = -5