Question

If \( A = \begin{bmatrix} 4 & 5 2 & 1 \end{bmatrix} \), find \( A^{-1} \).

• A. \( \begin{bmatrix} 1/6 & -5/6 -1/3 & 2/3 \end{bmatrix} \)
• B. \( \begin{bmatrix} -1/6 & 5/6 1/3 & -2/3 \end{bmatrix} \)
• C. \( \begin{bmatrix} -1 & 5 2 & -4 \end{bmatrix} \)
• D. \( \begin{bmatrix} 4 & -5 -2 & 1 \end{bmatrix} \)

Hint:

For \(2 \times 2\) matrices, always use the shortcut: Swap diagonal elements, change signs of off-diagonal elements, then divide by determinant.

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 exists if \( |A| \neq 0 \), where \[ |A| = ad - bc \] The inverse is given by: \[ A^{-1} = \frac{1}{|A|} \, \text{adj}(A) \] For a \(2 \times 2\) matrix: \[ \text{adj}(A) = \begin{bmatrix} d & -b -c & a \end{bmatrix} \]

Step 1: Finding determinant.
Given: \[ A = \begin{bmatrix} 4 & 5 2 & 1 \end{bmatrix} \] \[ |A| = (4 \times 1) - (5 \times 2) \] \[ |A| = 4 - 10 = -6 \] Since \( |A| \neq 0 \), inverse exists.

Step 2: Finding adjoint of \(A\).
Swap diagonal elements and change signs of off-diagonal elements: \[ \text{adj}(A) = \begin{bmatrix} 1 & -5 -2 & 4 \end{bmatrix} \]

Step 3: Applying inverse formula.
\[ A^{-1} = \frac{1}{-6} \begin{bmatrix} 1 & -5 -2 & 4 \end{bmatrix} \] Multiply each entry: \[ A^{-1} = \begin{bmatrix} -1/6 & 5/6 2/6 & -4/6 \end{bmatrix} \] Simplify: \[ A^{-1} = \begin{bmatrix} -1/6 & 5/6 1/3 & -2/3 \end{bmatrix} \]