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

To find the value of \(x\) in the matrix \(A\), we need to use the property of inverse matrices. Given:

\(A = \begin{bmatrix} x & 3 \\ 2 & 4 \end{bmatrix}\)

and

\(A^{-1} = \begin{bmatrix} -2 & 1.5 \\ 1 & -0.5 \end{bmatrix}\)

The product of a matrix and its inverse is the identity matrix. Therefore:

\(A \cdot A^{-1} = I\)

\(\begin{bmatrix} x & 3 \\ 2 & 4 \end{bmatrix} \cdot \begin{bmatrix} -2 & 1.5 \\ 1 & -0.5 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

Let's multiply the matrices on the left-hand side:

• First row, first column: \(x \cdot (-2) + 3 \cdot 1 = -2x + 3\)
• First row, second column: \(x \cdot 1.5 + 3 \cdot (-0.5) = 1.5x - 1.5\)
• Second row, first column: \(2 \cdot (-2) + 4 \cdot 1 = -4 + 4 = 0\)
• Second row, second column: \(2 \cdot 1.5 + 4 \cdot (-0.5) = 3 - 2 = 1\)

The equality \(-2x + 3 = 1\) arises from the first element of the identity matrix, solving gives:

\[-2x + 3 = 1\]

\[-2x = 1 - 3\]

\[-2x = -2\]

\[x = 1\]

Therefore, the correct value of \(x\) is 1.