Question

If $$ A = \begin{bmatrix} 2 & 3 & 4 \\0 & 4 & 2 \\0 & 0 & 3 \end{bmatrix} $$ then the eigenvalues of $ \text{adj}(A) $ are

• A. 12, 5, 3
• B. 12, 6, 8
• C. 10, 2, 3
• D. 2, 3, 4

Hint:

The eigenvalues of the adjugate matrix are obtained by multiplying the determinant of the matrix with the reciprocals of the eigenvalues of the original matrix.

Answer 1

The Correct Option is B

We are given the matrix \( A \). The eigenvalues of the adjugate (or adjoint) matrix \( \text{adj}(A) \) can be found from the properties of determinants. The adjugate of a matrix \( A \) has the following relationship with the determinant of \( A \): \[ \text{adj}(A) = \text{det}(A) \cdot A^{-1} \] From the matrix \( A \), we can first compute the determinant: \[ \text{det}(A) = 2 \cdot \begin{vmatrix} 4 & 2 \\0 & 3 \end{vmatrix} = 2 \cdot (4 \cdot 3 - 2 \cdot 0) = 2 \cdot 12 = 24 \] Now, for a \( 3 \times 3 \) matrix, the eigenvalues of the adjugate matrix are related to the eigenvalues of the original matrix by the following property: If \( \lambda_1, \lambda_2, \lambda_3 \) are the eigenvalues of \( A \), then the eigenvalues of \( \text{adj}(A) \) are given by: \[ \text{Eigenvalues of adj}(A) = \text{det}(A) \times \frac{1}{\lambda_1}, \text{det}(A) \times \frac{1}{\lambda_2}, \text{det}(A) \times \frac{1}{\lambda_3} \] The eigenvalues of \( A \) are the diagonal elements of the upper triangular matrix \( A \), which are 2, 4, and 
(3) Therefore, the eigenvalues of \( \text{adj}(A) \) are: \[ \text{det}(A) \times \frac{1}{2}, \text{det}(A) \times \frac{1}{4}, \text{det}(A) \times \frac{1}{3} \] Substituting the determinant \( \text{det}(A) = 24 \), we get: \[ 24 \times \frac{1}{2} = 12, \quad 24 \times \frac{1}{4} = 6, \quad 24 \times \frac{1}{3} = 8 \] Thus, the eigenvalues of \( \text{adj}(A) \) are \( 12, 6, 8 \). 
Conclusion: The eigenvalues of \( \text{adj}(A) \) are 12, 6, 8.