Question:

The eigenvalues of the matrix \( A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \) are:

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For 2x2 matrices, the eigenvalues can be found by solving the characteristic equation \( {det}(A - \lambda I) = 0 \), which leads to a quadratic equation.
Updated On: May 2, 2025
  • \( \pm \sqrt{-1} \)
  • -1 and 1
  • \( 1 + \sqrt{-1} \) and \( 1 - \sqrt{-1} \)
  • \( -1 + \sqrt{-1} \) and \( -1 - \sqrt{-1} \)
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The Correct Option is A

Solution and Explanation

We are given the matrix \( A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \). To find the eigenvalues, we use the characteristic equation:
\[ \text{det}(A - \lambda I) = 0 \]
where \( I \) is the identity matrix and \( \lambda \) represents the eigenvalues. Substituting the values:
\[ \text{det} \begin{bmatrix} 0 - \lambda & -1 \\ 1 & 0 - \lambda \end{bmatrix} = 0 \]
This simplifies to:
\[ \lambda^2 + 1 = 0 \]
Solving for \( \lambda \), we get:
\[ \lambda = \pm \sqrt{-1} \]
Thus, the eigenvalues are \( \pm \sqrt{-1} \), which are imaginary numbers, corresponding to option (A).
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