Find the inverse of {pmatrix} 3 & 5 \\ 1 & 2 {pmatrix}

Correct Answer is : (2−5−13){pmatrix} 2 & -5 \\ -1 & 3 {pmatrix}(2−1−53)

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

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

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

{pmatrix} 5 & 1 \\ 2 & 3 {pmatrix}

Correct Answer is : (2−5−13){pmatrix} 2 & -5 \\ -1 & 3 {pmatrix}(2−1−53)

Test Index

WAEC Further Mathematics 2020 Paper

Question : 2

Total: 39

Find the inverse of

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

Solution:

Concept:

For a

2 \times 2

matrix

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

, the inverse is

\dfrac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

, provided

ad-bc \neq 0

Explanation:

Let

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

Here,

a=3

b=5

c=1

d=2

Compute the determinant:

ad-bc = (3)(2) - (5)(1) = 6-5 = 1

Since the determinant is

1 \neq 0

, the inverse exists.

Using the inverse formula:

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

Therefore,

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

Answer:

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

This is option B.

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