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).