Question

The eigenvalues of a $3 \times 3$ matrix A are 1, 3, 7 then the eigenvalues of $\text{adj}(A)$ are

• A. \( 1, \frac{3}{4}, \frac{7}{4} \)
• B. \( \frac{1}{21}, \frac{3}{21}, \frac{7}{21} \)
• C. \( \frac{1}{21}, \frac{3}{21}, \frac{21}{7} \)
• D. \( 21, 7, 3 \)

Hint:

For an $n \times n$ matrix A with eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n$, the eigenvalues of $\text{adj}(A)$ are $\frac{\det(A)}{\lambda_1}, \frac{\det(A)}{\lambda_2}, \ldots, \frac{\det(A)}{\lambda_n}$. This formula is derived from the property $A \cdot \text{adj}(A) = \det(A) \cdot I$, where $I$ is the identity matrix. If $\mathbf{v}$ is an eigenvector of A with eigenvalue $\lambda$, then $A\mathbf{v} = \lambda\mathbf{v}$. Multiplying by $\text{adj}(A)$ gives $\text{adj}(A) A\mathbf{v} = \text{adj}(A) \lambda\mathbf{v}$, which simplifies to $\det(A) I \mathbf{v} = \det(A) \mathbf{v} = \text{adj}(A) \lambda\mathbf{v}$. Thus, $\text{adj}(A)\mathbf{v} = \frac{\det(A)}{\lambda}\mathbf{v}$.

Answer 1

The Correct Option is D

Let the eigenvalues of a matrix A be \( \lambda_1, \lambda_2, \ldots, \lambda_n \). The product of the eigenvalues is equal to the determinant of the matrix, i.e., \( \det(A) = \lambda_1 \cdot \lambda_2 \cdot \ldots \cdot \lambda_n \). The eigenvalues of \( \text{adj}(A) \) are given by \( \frac{\det(A)}{\lambda_i} \) for each eigenvalue \( \lambda_i \) of A. Given that the eigenvalues of the \( 3 \times 3 \) matrix A are \( \lambda_1 = 1 \), \( \lambda_2 = 3 \), and \( \lambda_3 = 7 \).

Step 1: Calculate the determinant of A.
\[ \det(A) = \lambda_1 \cdot \lambda_2 \cdot \lambda_3 = 1 \cdot 3 \cdot 7 = 21 \]

Step 2: Calculate the eigenvalues of \( \text{adj}(A) \)
The eigenvalues of \( \text{adj}(A) \) are \( \frac{\det(A)}{\lambda_1} \), \( \frac{\det(A)}{\lambda_2} \), and \( \frac{\det(A)}{\lambda_3} \).

For \( \lambda_1 = 1 \):
\[ \text{Eigenvalue}_1 = \frac{21}{1} = 21 \]

For \( \lambda_2 = 3 \):
\[ \text{Eigenvalue}_2 = \frac{21}{3} = 7 \]

For \( \lambda_3 = 7 \):
\[ \text{Eigenvalue}_3 = \frac{21}{7} = 3 \]

Therefore, the eigenvalues of \( \text{adj}(A) \) are 21, 7, and 3.