91Ó°ÊÓ

Identify the given matrix. For the first matrix: \[ A = \begin{pmatrix} 0 & -1-k^{2} \ 1 & 2 \end{pmatrix} \] and for the second matrix: \[ B = \begin{pmatrix} 0 & -1-k^{2} \ 1 & 2k \end{pmatrix} \]

The characteristic equation of a matrix \(C = \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is given by the determinant of \( C - \lambda I \), where \( \lambda \) is an eigenvalue and \( I \) is the identity matrix. Thus, for matrix \(A\): \[ \text{det}(A - \lambda I) = \begin{vmatrix} 0 - \lambda & -1-k^{2} \ 1 & 2 - \lambda \end{vmatrix} \]For matrix \(B\):\[ \text{det}(B - \lambda I) = \begin{vmatrix} 0 - \lambda & -1-k^{2} \ 1 & 2k - \lambda \end{vmatrix} \]

Calculate the determinant for both matrices:For matrix \(A\):\[ \text{det}(A - \lambda I) = (-\lambda)(2 - \lambda) - (-1-k^{2})(1) \]Simplify the expression:\[ -2\lambda + \lambda^{2} + 1 + k^{2} = \lambda^{2} - 2\lambda + 1 + k^{2} \]For matrix \(B\):\[ \text{det}(B - \lambda I) = (-\lambda)(2k - \lambda) - (-1-k^{2})(1) \]Simplify the expression:\[ -2k\lambda + \lambda^{2} + 1 + k^{2} = \lambda^{2} - 2k\lambda + 1 + k^{2} \]

Solve the quadratic equations for \( \lambda \). For matrix \(A\):\[ \lambda^{2} - 2\lambda + 1 + k^{2} = 0 \]For matrix \(B\):\[ \lambda^{2} - 2k\lambda + 1 + k^{2} = 0 \]Using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \) where \(a = 1\), \(b = -2\) and \(c = 1 + k^{2}\) for matrix \(A\), and \(a = 1\), \(b = -2k\) and \(c = 1 + k^{2}\) for matrix \(B\):For matrix \(A\):\[ \lambda = \frac{2 \pm \sqrt{4 - 4(1 + k^{2})}}{2} = \frac{2 \pm \sqrt{4 - 4 - 4k^{2}}}{2} = \frac{2 \pm \sqrt{-4k^{2}}}{2} \]\[ \lambda = \frac{2 \pm 2i|k|}{2} = 1 \pm i|k| \]For matrix \(B\):\[ \lambda = \frac{2k \pm \sqrt{4k^{2} - 4(1 + k^{2})}}{2} = \frac{2k \pm \sqrt{4k^{2} - 4 - 4k^{2}}}{2} = \frac{2k \pm \sqrt{-4}}{2} \]\[ \lambda = \frac{2k \pm 2i}{2} = k \pm i \]

Compare the eigenvalues obtained.For the first matrix, the eigenvalues are \( 1 + i|k| \) and \( 1 - i|k| \).For the second matrix, the eigenvalues are \( k + i \) and \( k - i \).Notice that as \( k \) changes over \([-\infty, \infty]\):- For the first matrix, the eigenvalues rotate around the point 1 on the real axis with an imaginary part that scales linearly with \(|k|\).- For the second matrix, the real part of the eigenvalues scales linearly with \(k\), whereas the imaginary part remains constant at \( \pm i \).