Question

The Trace and Determinant of a 2 × 2 nonsingular matrix A are 12 and 32, respectively. The eigen values of \(A^{-1}\) are \(\underline{\hspace{2cm}}\) and \(\underline{\hspace{2cm}}\)

• A. 0.6, 0.8
• B. 0.25, 0.125
• C. 6, 16
• D. 1/12, 1/32

Hint:

The eigenvalues of \(A^{-1}\) are the reciprocals of the eigenvalues of \(A\).

Answer 1

The Correct Option is B

Given the trace and determinant of matrix \(A\): - The trace of a matrix is the sum of its eigenvalues: \[ \text{Tr}(A) = \lambda_1 + \lambda_2 = 12. \] - The determinant of a matrix is the product of its eigenvalues: \[ \text{Det}(A) = \lambda_1 \lambda_2 = 32. \] The eigenvalues of \(A^{-1}\) are the reciprocals of the eigenvalues of \(A\). Let the eigenvalues of \(A\) be \(\lambda_1\) and \(\lambda_2\). The eigenvalues of \(A^{-1}\) are \(\frac{1}{\lambda_1}\) and \(\frac{1}{\lambda_2}\). We know that: \[ \lambda_1 + \lambda_2 = 12, \lambda_1 \lambda_2 = 32. \] Using the quadratic formula, we solve for the eigenvalues of \(A\): \[ \lambda_1, \lambda_2 = \frac{12 \pm \sqrt{12^2 - 4(32)}}{2} = \frac{12 \pm \sqrt{144 - 128}}{2} = \frac{12 \pm \sqrt{16}}{2} = \frac{12 \pm 4}{2}. \] Thus, the eigenvalues of \(A\) are: \[ \lambda_1 = 8, \lambda_2 = 4. \] The eigenvalues of \(A^{-1}\) are the reciprocals: \[ \frac{1}{\lambda_1} = \frac{1}{8}, \frac{1}{\lambda_2} = \frac{1}{4}. \] Therefore, the eigenvalues of \(A^{-1}\) are Option (B): \( 0.25 \) and \( 0.125 \).