Question

The smallest eigenvalue and the corresponding eigenvector of the matrix \( \begin{bmatrix} 2 & -2 \\ -1 & 6 \end{bmatrix} \) respectively are 

• A. \( 1.55 \) and \( \begin{bmatrix} 2.00 \\ 0.45 \end{bmatrix} \)
• B. \( 2.00 \) and \( \begin{bmatrix} 1.00 \\ 1.00 \end{bmatrix} \)
• C. \( 1.55 \) and \( \begin{bmatrix} -2.55  \\ -0.45 \end{bmatrix} \)
• D. \( 1.55 \) and \( \begin{bmatrix} 2.00  \\ -0.45 \end{bmatrix} \)

Hint:

To find eigenvalues, solve the characteristic equation \( \text{det}(A - \lambda I) = 0 \). Once you find the eigenvalues, substitute them into \( (A - \lambda I) v = 0 \) to find the corresponding eigenvectors.

Answer 1

The Correct Option is A

To find the eigenvalues and eigenvectors of the given matrix: \[ A = \begin{bmatrix} 2 & -2 \\ -1 & 6 \end{bmatrix} \] We solve the characteristic equation \( \text{det}(A - \lambda I) = 0 \), where \( \lambda \) is the eigenvalue and \( I \) is the identity matrix. First, subtract \( \lambda I \) from \( A \): \[ A - \lambda I = \begin{bmatrix} 2 - \lambda & -2 \\ -1 & 6 - \lambda \end{bmatrix} \] Now, compute the determinant: \[ \text{det}(A - \lambda I) = (2 - \lambda)(6 - \lambda) - (-2)(-1) \] \[ = (2 - \lambda)(6 - \lambda) - 2 \] Expanding the equation: \[ = 12 - 2\lambda - 6\lambda + \lambda^2 - 2 \] \[ = \lambda^2 - 8\lambda + 10 = 0 \] Solving this quadratic equation: \[ \lambda = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(10)}}{2(1)} \] \[ \lambda = \frac{8 \pm \sqrt{64 - 40}}{2} \] \[ \lambda = \frac{8 \pm \sqrt{24}}{2} = \frac{8 \pm 4.9}{2} \] \[ \lambda_1 = 1.55 \text{(smallest eigenvalue)} \lambda_2 = 6.45 \] Next, we substitute \( \lambda_1 = 1.55 \) into \( (A - \lambda I) v = 0 \) to find the corresponding eigenvector: \[ \begin{bmatrix} 2 - 1.55 & -2 \\ -1 & 6 - 1.55 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = 0 \] \[ \begin{bmatrix} 0.45 & -2 \\ -1 & 4.45 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = 0 \] Solving this system of linear equations yields: \[ v_1 = 2, v_2 = 0.45 \] Thus, the corresponding eigenvector is \( \begin{bmatrix} 2.00 \\ 0.45 \end{bmatrix} \). Therefore, the correct answer is option (A).
Final Answer: (A) \( 1.55 \) and \( \begin{bmatrix} 2.00 \\ 0.45 \end{bmatrix} \)