Question

Given that $ A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix} $, matrix $ A $ is:

• A. \[ 7 \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix} \]
• B. \[ \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix} \]
• C. \[ \frac{1}{7} \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix} \]
• D. \[ \frac{1}{49} \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix} \]

Hint:

To find \( A \) from \( A^-1 \), multiply the inverse by the scalar reciprocal.

Answer 1

The Correct Option is A

Given that \( A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix} \), we need to find matrix \( A \).

The inverse of a \( 2 \times 2 \) matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by: \[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] Applying this to \( A^{-1} \): \[ A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix} \] The determinant of \( A^{-1} \) is: \[ \det(A^{-1}) = \left(\frac{1}{7}\right)^2 (2 \cdot 2 - 1 \cdot (-3)) = \frac{1}{49} (4 + 3) = \frac{7}{49} = \frac{1}{7} \] Now, the inverse of \( A^{-1} \) (which is \( A \)) is: \[ A = \frac{1}{\det(A^{-1})} \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix} = 7 \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix} \] Therefore, the correct answer is: \[ \boxed{A = 7 \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix}} \]