Given \(\begin{vmatrix} 2 & -3 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} -6 \\ k \end{vmatrix}
FURTHER MATHEMATICS
WAEC 2022
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: B. -5
Explanation
\(\begin{vmatrix} 2 & -3 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} -6 \\ k \end{vmatrix} \begin{vmatrix} 3 \\ -26 \end{vmatrix} = 15\)
\(\begin{vmatrix} 2[-6] & - 3k \\ 1[-6] & + 4k \end{vmatrix} = \begin{vmatrix} 3 \\ -26 \end{vmatrix}\)
\(\begin{vmatrix} -12 & - 3k \\ -6 & + 4k \end{vmatrix} = \begin{vmatrix} 3 \\ -26 \end{vmatrix}\)
-12 - 3k = 3
-3k = 3 + 12
k = \(\frac{15}{-3}\)
k = -5
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