If B={pmatrix} 2 & 5 \\ 1 & 3 {pmatrix}, find B^{-1}.

Correct Answer is : A=(3−5−12)A={pmatrix} 3 & -5 \\ -1 & 2 {pmatrix}A=(3−1−52)

A={pmatrix} -3 & -5 \\ 1 & 2 {pmatrix}

A={pmatrix} 3 & -5 \\ 1 & 2 {pmatrix}

A={pmatrix} 3 & -5 \\ -1 & 2 {pmatrix}

A={pmatrix} -3 & 5 \\ 1 & -2 {pmatrix}

Correct Answer is : A=(3−5−12)A={pmatrix} 3 & -5 \\ -1 & 2 {pmatrix}A=(3−1−52)

Test Index

WAEC Further Mathematics 2017 Paper

Question : 17

Total: 40

B=\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}

B^{-1}

Solution:

Concept:

For a

2 \times 2

matrix, the inverse is found using the formula

B^{-1} = \frac{1}{\det(B)} \times \text{adj}(B)

Explanation:

Let

B = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}

First, find the determinant of

B

\det(B) = (2)(3) - (5)(1) = 6 - 5 = 1

Since

\det(B) \neq 0

, the inverse exists.

For a matrix

\begin{pmatrix} p & q \\ r & s \end{pmatrix}

, the inverse is

\frac{1}{ps - qr} \begin{pmatrix} s & -q \\ -r & p \end{pmatrix}

Here,

p = 2

q = 5

r = 1

, and

s = 3

B^{-1} = \frac{1}{1} \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}

Therefore,

B^{-1} = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}

This matches option C.

Answer:

B^{-1} = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}

, which is Option C.

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