Question

Eigenvalues of the matrix $\begin{bmatrix}0 & 1 \\ -2 & 3 \end{bmatrix}$ are

• A. 1 and 2
• B. 1 and 3
• C. 1 and $-2$
• D. 0 and 3

Hint:

For a $2 \times 2$ matrix, the eigenvalues are roots of $\lambda^2 - (\text{trace})\lambda + (\det) = 0$.

Answer 1

The Correct Option is D

Let
\[ A = \begin{bmatrix} 0 & 1 \\ -2 & 3 \end{bmatrix}. \]
Eigenvalues satisfy
\[ \det(A - \lambda I) = 0. \]
Compute the determinant:
\[ \det \begin{bmatrix} -\lambda & 1 \\ -2 & 3 - \lambda \end{bmatrix} = (-\lambda)(3 - \lambda) - (1)(-2). \]
\[ = -3\lambda + \lambda^2 + 2 = \lambda^2 - 3\lambda + 2. \]
Solve the quadratic:
\[ \lambda^2 - 3\lambda + 2 = 0. \]
\[ (\lambda - 1)(\lambda - 2) = 0. \]
Thus, eigenvalues are
\[ \lambda = 1,\quad \lambda = 2. \]
Final Answer: (A)