Question

If \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \), find the inverse of matrix \( A \).

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

Hint:

To find the inverse of a 2x2 matrix, compute the determinant \( ad - bc \), then use \( A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \).

Answer 1

The Correct Option is A

For a 2x2 matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the inverse is: \[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] Given \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \), where \( a = 1 \), \( b = 2 \), \( c = 3 \), \( d = 4 \): - Determinant: \( ad - bc = 1 \cdot 4 - 2 \cdot 3 = 4 - 6 = -2 \) - Inverse: \[ A^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} \frac{4}{-2} & \frac{-2}{-2} \\ \frac{-3}{-2} & \frac{1}{-2} \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix} \] Thus, the inverse is: \[ \boxed{\begin{bmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}} \]