Question

If \[ A=\begin{bmatrix} x & 3 \\ 2 & 4 \end{bmatrix} \quad \text{and} \quad A^{-1}=\begin{bmatrix} -2 & 1.5 \\ 1 & -0.5 \end{bmatrix}, \] then the value of \(x\) is

• A. \(-2\)
• B. \(1\)
• C. \(1.5\)
• D. \(-0.5\)

Hint:

For inverse matrix questions, use the standard inverse formula of a \(2 \times 2\) matrix and compare corresponding entries.

Answer 1

The Correct Option is B

Concept:
For a \(2 \times 2\) matrix: \[ A=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \] the inverse is: \[ A^{-1}=\frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]

Step 1: Here, \[ A=\begin{bmatrix} x & 3 \\ 2 & 4 \end{bmatrix} \]

Step 2: The determinant of \(A\) is: \[ |A|=4x-6 \]

Step 3: Therefore, \[ A^{-1}=\frac{1}{4x-6} \begin{bmatrix} 4 & -3 \\ -2 & x \end{bmatrix} \]

Step 4: Given: \[ A^{-1}=\begin{bmatrix} -2 & 1.5 \\ 1 & -0.5 \end{bmatrix} \]

Step 5: Compare the first element: \[ \frac{4}{4x-6}=-2 \]

Step 6: Solve: \[ 4=-2(4x-6) \] \[ 4=-8x+12 \] \[ 8x=8 \] \[ x=1 \] \[ \boxed{1} \]