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WAEC Further Mathematics 2019 Paper

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Question : 15

Total: 39

If

P=\begin{pmatrix}2 & 3\\ -4 & 1\end{pmatrix}, Q=\binom{6}{8}

P Q = k\binom{45}{-20}

. Find the value of

k

.

- $\;\frac{5}{4}$
- $\;\frac{4}{5}$
$\;\frac{4}{5}$
$\;\frac{5}{4}$

Solution:

Concept:

Multiply the matrix

P

by the column vector

Q

, then compare the resulting vector with the given expression to solve for

k

.

Explanation:

First compute

PQ

by multiplying the

2 \times 2

matrix with the

2 \times 1

column vector.

PQ = \begin{pmatrix}2 & 3\\ -4 & 1\end{pmatrix} \binom{6}{8}

Row 1:

(2)(6) + (3)(8) = 12 + 24 = 36

Row 2:

(-4)(6) + (1)(8) = -24 + 8 = -16

Thus,

PQ = \binom{36}{-16}

.

Given

PQ = k\binom{45}{-20}

, so

\binom{36}{-16} = k\binom{45}{-20}

.

Compare the first components:

36 = 45k

.

Hence,

k = \frac{36}{45} = \frac{4}{5}

.

Check with the second component:

-16 = k(-20)

, so

k = \frac{-16}{-20} = \frac{4}{5}

, which matches.

Answer:

C.

\frac{4}{5}

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