Question

Let \( A = \begin{pmatrix} a & 0 \\ c & d \end{pmatrix} \) be a real matrix, where \( ad = 1 \) and \( c \ne 0 \). If \( A^{-1} + (\text{adj} \, A)^{-1} = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \), then \( (\alpha, \beta, \gamma, \delta) \) is equal to

• A. \( (a + d, 0, 0, a + d) \)
• B. \( (a + d, 0, c, a + d) \)
• C. \( (a, 0, 0, d) \)
• D. \( (a, 0, c, d) \)

Answer 1

The Correct Option is A

1. Matrix Inverse and Adjoint: - The inverse of \( A \) is given by: \[ A^{-1} = \frac{1}{\det A} \cdot \text{adj } A = \text{adj } A, \] since \( \det A = ad - 0 = 1 \). 

2. Simplify the Given Expression: - Substituting \( A^{-1} = \text{adj } A \): \[ A^{-1} + (\text{adj } A)^{-1} = \text{adj } A + (\text{adj } A)^{-1}. \] - Since \( (\text{adj } A)^{-1} = A \) (property of adjugate matrices), we get: \[ A^{-1} + (\text{adj } A)^{-1} = \text{adj } A + A. \] 3. Calculate \((\alpha, \beta, \gamma, \delta)\): - Adding \( A \) and \( \text{adj } A \): \[ \begin{pmatrix} a & 0 \\ c & d \end{pmatrix} + \begin{pmatrix} d & 0 \\ -c & a \end{pmatrix} = \begin{pmatrix} a + d & 0 \\ 0 & a + d \end{pmatrix}. \] 4. Conclusion: - The resulting matrix is: \[ \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} = \begin{pmatrix} a + d & 0 \\ 0 & a + d \end{pmatrix}. \]