Question

If \[ A = \begin{bmatrix} 2 & -3 \\ 4 & 6 \end{bmatrix} \] then \[ A^{-1} = \] The correct answer is:

• A. \( \begin{bmatrix} 4 & 6 \\ 8 & 12 \end{bmatrix} \)
• B. \( \begin{bmatrix} 4 & -6 \\ 8 & 12 \end{bmatrix} \)
• C. \( \begin{bmatrix} 4 & 6 \\ 8 & 12 \end{bmatrix} \)
• D. \( \begin{bmatrix} 4 & -6 \\ 8 & 12 \end{bmatrix} \)

Hint:

For a 2x2 matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the inverse is given by: \[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}. \] Always check the determinant before finding the inverse. If the determinant is zero, the matrix is not invertible.

Answer 1

The Correct Option is B

For a 2×2 matrix \[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, \] the inverse is given by: \[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}. \]

Here, \( a = 2 \), \( b = -3 \), \( c = 4 \), \( d = 6 \).

Determinant: \[ \det(A) = (2)(6) - (-3)(4) = 12 + 12 = 24. \]

Now substitute into the formula: \[ A^{-1} = \frac{1}{24} \begin{bmatrix} 6 & 3 \\ -4 & 2 \end{bmatrix}. \]

Simplify each element: \[ A^{-1} = \begin{bmatrix} \frac{6}{24} & \frac{3}{24} \\ \frac{-4}{24} & \frac{2}{24} \end{bmatrix} = \begin{bmatrix} \frac{1}{4} & \frac{1}{8} \\ -\frac{1}{6} & \frac{1}{12} \end{bmatrix}. \]

Hence, the inverse of the given matrix is: \[ \boxed{ A^{-1} = \begin{bmatrix} \frac{1}{4} & \frac{1}{8} \\ -\frac{1}{6} & \frac{1}{12} \end{bmatrix}. } \]