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WAEC Further Mathematics 2013 Paper

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Question : 15

Total: 40

Given that

P = \begin{pmatrix} y-2 & y-1 \\ y-4 & y+2 \end{pmatrix}

\left| PI \right| = -23

, find the value of y .

-4
-3
-1
2

Solution:

Concept:

For a matrix

P = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

, the determinant is

ad - bc

.

Explanation:

Given

P = \begin{pmatrix} y-2 & y-1 \\ y-4 & y+2 \end{pmatrix}

and

|P| = -23

.

First, find

|P|

:

|P| = (y-2)(y+2) - (y-1)(y-4)

= y^2 - 4 - (y^2 - 5y + 4)

= y^2 - 4 - y^2 + 5y - 4

= 5y - 8

Set the determinant equal to

-23

:

5y - 8 = -23

5y = -23 + 8 = -15

y = \frac{-15}{5} = -3

Answer:

y = -3

(Option B).

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