Given the matrix A = {vmatrix} 3 & -2 \\ 1 & 6 {vmatrix}. Find the inverse of matrix A.

Correct Answer is : ∣310110−120320∣{vmatrix} 3/10 & 1/10 \\ -1/20 & 3/20 {vmatrix}10320−1101203

{vmatrix} 6 & 2 \\ 1 & 6 {vmatrix}

{vmatrix} 2/11 & 1/12 \\ 3/20 & 1/10 {vmatrix}

{vmatrix} -3 & 2 \\ -1 & -6 {vmatrix}

{vmatrix} 3/10 & 1/10 \\ -1/20 & 3/20 {vmatrix}

Correct Answer is : ∣310110−120320∣{vmatrix} 3/10 & 1/10 \\ -1/20 & 3/20 {vmatrix}10320−1101203

Test Index

JAMB Mathematics 2019 Paper

Question : 70

Total: 73

Given the matrix

A = \begin{vmatrix} 3 & -2 \\ 1 & 6 \end{vmatrix}

. Find the inverse of matrix A.

$\begin{vmatrix} 6 & 2 \\ 1 & 6 \end{vmatrix}$
$\begin{vmatrix} \frac{2}{11} & \frac{1}{12} \\ \frac{3}{20} & \frac{1}{10} \end{vmatrix}$
$\begin{vmatrix} -3 & 2 \\ -1 & -6 \end{vmatrix}$
$\begin{vmatrix} \frac{3}{10} & \frac{1}{10} \\ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}$

Solution:

Concept:

The inverse of a

2 \times 2

matrix is found by dividing the adjugate matrix by the determinant.

Explanation:

For

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

, the inverse is

A^{-1} = \frac{1}{ad - bc} \begin{vmatrix} d & -b \\ -c & a \end{vmatrix}

Given

A = \begin{vmatrix} 3 & -2 \\ 1 & 6 \end{vmatrix}

, find the determinant:

|A| = (3)(6) - (-2)(1) = 18 + 2 = 20

The adjugate matrix is

\begin{vmatrix} 6 & 2 \\ -1 & 3 \end{vmatrix}

Therefore,

A^{-1} = \frac{1}{20} \begin{vmatrix} 6 & 2 \\ -1 & 3 \end{vmatrix}

Simplify each entry:

A^{-1} = \begin{vmatrix} \frac{6}{20} & \frac{2}{20} \\ \frac{-1}{20} & \frac{3}{20} \end{vmatrix} = \begin{vmatrix} \frac{3}{10} & \frac{1}{10} \\ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}

Answer:

Option D:

\begin{vmatrix} \frac{3}{10} & \frac{1}{10} \\ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}

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