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: {Matrix Inversion Property}
If the inverse of a matrix \( A^{-1} \) is provided, the original matrix \( A \) can be obtained by taking the reciprocal of the scalar multiple associated with \( A^{-1} \).
Step 2: {Calculate \( A \)}
Given \( A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix}\), we calculate \( A \) by multiplying \( A^{-1} \) by 7:\( A = 7 \times A^{-1} = 7 \cdot \frac{1}{7} \begin{bmatrix} 2 & 1\\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix}. \)
Step 3: {Verify the options}
The calculated matrix \( A \) corresponds to option (B).