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WAEC Further Mathematics 2018 Paper

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Question : 15

Total: 40

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

k>0

.

1
2
3
4

Solution:

Concept:

Determinant of a

2 \times 2

matrix and solving a quadratic equation.

Explanation:

For a matrix

\begin{vmatrix} a & b \\ c & d \end{vmatrix}

, the determinant is

ad - bc

.

First determinant:

\begin{vmatrix} k & k \\ 4 & k \end{vmatrix} = (k)(k) - (4)(k) = k^2 - 4k

.

Second determinant:

\begin{vmatrix} 2 & 3 \\ -1 & k \end{vmatrix} = (2)(k) - (3)(-1) = 2k + 3

.

Add them:

(k^2 - 4k) + (2k + 3) = 6

.

Simplify:

k^2 - 2k + 3 = 6

, so

k^2 - 2k - 3 = 0

.

Factorise:

(k + 1)(k - 3) = 0

.

Thus,

k = -1

or

k = 3

.

Given

k > 0

, the valid value is

k = 3

.

Answer:

k = 3

(Option C).

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