91Ó°ÊÓ

To find the eigenvalues and eigenvectors, first express the transformation in matrix form. The transformation given is \( T \): \[ T\begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} = \begin{bmatrix} x_1 \ -5x_1 + 3x_2 - 5x_3 \ -3x_1 - 2x_2 \end{bmatrix} \]This transformation corresponds to the matrix:\[ A = \begin{bmatrix} 1 & 0 & 0 \ -5 & 3 & -5 \ -3 & -2 & 0 \end{bmatrix} \]

The eigenvalues \( \lambda \) of a matrix \( A \) are the solutions to the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. Calculate \( A - \lambda I \):\[ A - \lambda I = \begin{bmatrix} 1 - \lambda & 0 & 0 \ -5 & 3 - \lambda & -5 \ -3 & -2 & -\lambda \end{bmatrix} \]

To solve \( \det(A - \lambda I) = 0 \), calculate the determinant of the matrix:\[ \det(A - \lambda I) = (1 - \lambda) \cdot ((3 - \lambda)(-\lambda) - (-5)(-2)) \]simplify the expression:\[ = (1 - \lambda)((3-\lambda)(-\lambda) - 10) \]\[ = (1 - \lambda)(-3\lambda + \lambda^2 - 10) \]\[ = (1 - \lambda)(\lambda^2 - 3\lambda - 10) \]

Set the expression equal to zero and solve for \( \lambda \):\[ (1 - \lambda)(\lambda^2 - 3\lambda - 10) = 0 \]The solutions to this equation are \( \lambda = 1 \) and the roots of the quadratic \( \lambda^2 - 3\lambda - 10 = 0 \).

Use the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve \( \lambda^2 - 3\lambda - 10 = 0 \).\[ a = 1, \ b = -3, \ c = -10 \]\[ \Delta = b^2 - 4ac = 9 + 40 = 49 \]\[ \lambda = \frac{3 \pm \sqrt{49}}{2} = \frac{3 \pm 7}{2} \]Thus, \( \lambda = 5 \) and \( \lambda = -2 \).

For each eigenvalue, solve \((A - \lambda I)\mathbf{v} = 0\) to find the corresponding eigenvector.1. **\(\lambda = 1\):** \[ \begin{bmatrix} 0 & 0 & 0 \ -5 & 2 & -5 \ -3 & -2 & -1 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \ v_3 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \] Solving it gives the eigenvector \( \begin{bmatrix} 1 \ 5 \ -3 \end{bmatrix} \).2. **\(\lambda = 5\):** \[ \begin{bmatrix} -4 & 0 & 0 \ -5 & -2 & -5 \ -3 & -2 & -5 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \ v_3 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \] Solving it gives the eigenvector \( \begin{bmatrix} 0 \ 1 \ -2 \end{bmatrix} \).3. **\(\lambda = -2\):** \[ \begin{bmatrix} 3 & 0 & 0 \ -5 & 5 & -5 \ -3 & -2 & 2 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \ v_3 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \] Solving it gives the eigenvector \( \begin{bmatrix} 0 \ 1 \ 2 \end{bmatrix} \).