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.

Assessment: WAEC SSCE - Further Mathematics - 2022 Subject: Further Mathematics

Question 1 Report

-8

-5

-4

-3

Answer Details

$| \begin{matrix} 2 & - 3 \\ 1 & 4 \end{matrix} | | \begin{matrix} - 6 \\ k \end{matrix} | | \begin{matrix} 3 \\ - 26 \end{matrix} | = 15$

$| \begin{matrix} 2 [- 6] & - 3 k \\ 1 [- 6] & + 4 k \end{matrix} | = | \begin{matrix} 3 \\ - 26 \end{matrix} |$

$| \begin{matrix} - 12 & - 3 k \\ - 6 & + 4 k \end{matrix} | = | \begin{matrix} 3 \\ - 26 \end{matrix} |$

-12 - 3k = 3

-3k = 3 + 12

k = $\frac{15}{- 3}$

k = -5

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.