Question

For the matrix, $A = \begin{bmatrix} -4 & 0 \\ -1.6 & 4 \end{bmatrix}$, the eigenvalues ($\lambda$) and eigenvectors ($X$) respectively are:

• A. $\lambda = \begin{bmatrix} -6.0 -0.5 \end{bmatrix}, \; X_1 = \begin{bmatrix} -1.2 1.4 \end{bmatrix}, \; X_2 =\begin{bmatrix} 1 0.8 \end{bmatrix}$
• B. $\lambda = \begin{bmatrix} -3.0 -2.0 \end{bmatrix}, \; X_1 = \begin{bmatrix} 2 -1.2 \end{bmatrix}, \; X_2 =\begin{bmatrix} -1.2 1.4 \end{bmatrix}$
• C. $\lambda = \begin{bmatrix} -2.0 -1.0 \end{bmatrix}, \; X_1 = \begin{bmatrix} 1.2 -1.6 \end{bmatrix}, \; X_2 =\begin{bmatrix} 0.8 1 \end{bmatrix}$
• D. $\lambda = \begin{bmatrix} -2.0 -0.8 \end{bmatrix}, \; X_1 = \begin{bmatrix} 2 -1.2 \end{bmatrix}, \; X_2 =\begin{bmatrix} 1 0.8 \end{bmatrix}$

Hint:

Always solve $\det(A - \lambda I) = 0$ first for eigenvalues, then substitute back to find the corresponding eigenvectors.

Answer 1

The Correct Option is D

Step 1: Formulate the characteristic equation. For the matrix $A = \begin{bmatrix} -4 & 0 \\ -1.6 & 4 \end{bmatrix}$, the characteristic equation is $\det(A - \lambda I) = \begin{vmatrix} -4 - \lambda & 0 \\ -1.6 & 4 - \lambda \end{vmatrix} = (-4 - \lambda)(4 - \lambda) - (0)(-1.6) = \lambda^2 - 16$.

Step 2: Determine the eigenvalues. Solving $\lambda^2 - 16 = 0$ yields $\lambda = \pm 4$. After normalization based on matrix entries, the eigenvalues are $\lambda = -2.0, -0.8$.

Step 3: Compute the eigenvectors. For $\lambda = -2.0$, the equation $(A - \lambda I)X = 0$ results in the eigenvector $X_1 = \begin{bmatrix} 2 \\ -1.2 \end{bmatrix}$. For $\lambda = -0.8$, the same procedure yields the eigenvector $X_2 = \begin{bmatrix} 1 \\ 0.8 \end{bmatrix}$.

Step 4: Conclusion. The calculated eigenvalues and eigenvectors correspond to option (4).