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
waec 2022furthermathematics

The correct answer is B. -5

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

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

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

-12 - 3k = 3

-3k = 3 + 12

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

k = -5

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