Question:

The eigenvalues of the matrix \( \begin{bmatrix} 2 & 1 - 2i \\ 1 + 2i & -2 \end{bmatrix} \) are ..........

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When dealing with complex elements in a matrix, use the identity \( (a - bi)(a + bi) = a^2 + b^2 \) to simplify the determinant.
Updated On: Jun 20, 2025
  • $-3i,\ 3i$
  • $-3,\ 3$
  • $-2,\ 2$
  • $-2i,\ 2i$
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The Correct Option is B

Solution and Explanation

To find the eigenvalues of matrix \( A = \begin{bmatrix} 2 & 1 - 2i \\ 1 + 2i & -2 \end{bmatrix} \), we solve the characteristic equation:

\[ \det(A - \lambda I) = 0 \]
\[ \Rightarrow \det\begin{bmatrix} 2 - \lambda & 1 - 2i \\ 1 + 2i & -2 - \lambda \end{bmatrix} = 0 \]
Using determinant formula:
\[ (2 - \lambda)(-2 - \lambda) - (1 - 2i)(1 + 2i) = 0 \]
\[ = -(2 - \lambda)(2 + \lambda) - [(1)^2 + (2)^2] = -[(\lambda^2 - 4)] - [1 + 4] = -(\lambda^2 - 4) - 5 \]
\[ = -\lambda^2 + 4 - 5 = -\lambda^2 - 1 \Rightarrow \lambda^2 + 1 = 0 \Rightarrow \lambda = \pm i \]
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