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 \( A^{-1} \) is given, the original matrix \( A \) is the reciprocal of the scalar multiple. 
Step 2: {Calculate \( A \)}
Given \( A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix}\) , we compute \( A \) by multiplying the inverse by \( 7 \), so:\( A = 7 \times A^{-1} = 7 \cdot \frac{1}{7} \begin{bmatrix} 2 & 1\\  -3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix}. \)

 Step 3: {Verify the options}
The correct matrix is option (B).