Question

If \[ A= \begin{bmatrix} 2 & 3\\ 1 & 4 \end{bmatrix} \] then \(A^{-1}=\)

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

Hint:

For: \[ \begin{bmatrix} a & b c & d \end{bmatrix} \] inverse is: \[ \frac{1}{ad-bc} \begin{bmatrix} d & -b -c & a \end{bmatrix} \] Interchange diagonal entries and change signs of off-diagonal entries.

Answer 1

The Correct Option is A

Concept: For matrix: \[ A= \begin{bmatrix} a & b\\ c & d \end{bmatrix} \] inverse is: \[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b\\ -c & a \end{bmatrix} \] provided determinant is non-zero.

Step 1: Finding determinant. \[ |A| = (2)(4)-(3)(1) \] \[ =8-3 \] \[ =5 \]

Step 2: Applying inverse formula. \[ A^{-1} = \frac{1}{5} \begin{bmatrix} 4 & -3\\ -1 & 2 \end{bmatrix} \] Final Answer: \[ \boxed{(A)\ \frac{1}{5}\begin{bmatrix}4 & -3-1 & 2\end{bmatrix}} \]