Question

The inverse of matrix \( A = \begin{bmatrix} 4 & -1 \\ 2 & 1 \end{bmatrix} \) is

• A. \( \frac{1}{6} \begin{bmatrix} 4 & 2 \\ -1 & -1 \end{bmatrix} \)
• B. \( \begin{bmatrix} \frac{1}{3} & \frac{1}{6} \\ \frac{2}{3} & -\frac{1}{6} \end{bmatrix} \)
• C. \( \begin{bmatrix} \frac{1}{6} & \frac{1}{6} \\ -\frac{2}{3} & \frac{2}{3} \end{bmatrix} \)
• D. \( \begin{bmatrix} -\frac{2}{3} & \frac{1}{6} \\ -\frac{1}{3} & -\frac{1}{6} \end{bmatrix} \)

Hint:

Always use the $2 \times 2$ inverse formula carefully and simplify fractions at the end.

Answer 1

The Correct Option is B

Given matrix \( A = \begin{bmatrix} 4 & -1 \\ 2 & 1 \end{bmatrix} \).
The formula for the inverse of a \( 2 \times 2 \) matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is:
\( A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \).
Here, \( a = 4 \), \( b = -1 \), \( c = 2 \), \( d = 1 \).
Determinant = \( 4 \times 1 - (-1) \times 2 = 4 + 2 = 6 \).
So, \( A^{-1} = \frac{1}{6} \begin{bmatrix} 1 & 1 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} \frac{1}{6} & \frac{1}{6} \\ -\frac{1}{3} & \frac{2}{3} \end{bmatrix} \).
Option (C) matches this inverse: \( \begin{bmatrix} \frac{1}{6} & \frac{1}{6} \\ -\frac{1}{3} & \frac{2}{3} \end{bmatrix} \).
So, option (C) is correct.