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

To find the inverse of a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, we use the formula:

where $\det(A) = (ad - bc)$ is the determinant of the matrix.

Step 1: Calculate the determinant of A

Given $A = \begin{pmatrix} 3 & -2 \\ 1 & 6 \end{pmatrix}$:

• $a = 3, b = -2, c = 1, d = 6$
• $\det(A) = (3 \times 6) - (-2 \times 1)$
• $\det(A) = 18 - (-2)$
• $\det(A) = 18 + 2 = 20$

Step 2: Apply the inverse formula

Now, swap the elements on the leading diagonal ($3$ and $6$) and change the signs of the elements on the other diagonal ($-2$ and $1$):

• The adjugate matrix is $\begin{pmatrix} 6 & 2 \\ -1 & 3 \end{pmatrix}$

Now, multiply by $\frac{1}{\det(A)}$:

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

Step 3: Simplify the fractions

• $\frac{6}{20} = \frac{3}{10}$
• $\frac{2}{20} = \frac{1}{10}$
• The matrix becomes $\begin{pmatrix} \frac{3}{10} & \frac{1}{10} \\ \frac{-1}{20} & \frac{3}{20} \end{pmatrix}$

Comparing this result to your options, it matches option D.

Exam Tip: Always remember the "Swap and Negate" rule for $2 \times 2$ matrices: Swap the positions of the top-left and bottom-right numbers, and change the signs of the top-right and bottom-left numbers. Don't forget to divide the final matrix by the determinant!