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 B

Step 1: Understanding Matrix Inverse Property
If the inverse of a matrix, \( A^{-1} \), is provided, the original matrix \( A \) can be found by taking the reciprocal of the scalar multiple applied to \( A^{-1} \).
Step 2: Deriving Matrix \( A \)
Given the inverse matrix \( A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix} \), we calculate \( A \) by multiplying \( A^{-1} \) by \( 7 \). This yields:
\[ A = 7 \times A^{-1} = 7 \cdot \frac{1}{7} \begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix}. \]
Step 3: Option Validation
The calculated matrix \( A \) corresponds to option (B).