Question

The imaginary parts of the eigenvalues of the matrix
\[ P = \begin{pmatrix} 3 & 2 & 5 \\ 2 & -3 & 6 \\ 0 & 0 & -3 \end{pmatrix} \] are

• A. 2, -2, 0
• B. 0, 0, 0
• C. 1, -1, 0
• D. 3, -3, 0

Hint:

When calculating eigenvalues, solving the characteristic equation \( \det(P - \lambda I) = 0 \) will give you the eigenvalues, and the imaginary part can be directly observed from the result.

Answer 1

The Correct Option is B

Step 1: Write the given matrix.
\[ P = \begin{pmatrix} 3 & 2 & 5 \\ 2 & -3 & 6 \\ 0 & 0 & -3 \end{pmatrix} \]

Step 2: Find the characteristic equation.
We calculate the determinant of \( (P - \lambda I) \):
\[ \begin{vmatrix} 3 - \lambda & 2 & 5 \\ 2 & -3 - \lambda & 6 \\ 0 & 0 & -3 - \lambda \end{vmatrix} = 0 \]

Since the last row has two zeros, expand along that row:
\[ (-3 - \lambda) \times \begin{vmatrix} 3 - \lambda & 2 \\ 2 & -3 - \lambda \end{vmatrix} = 0 \]

Step 3: Simplify the 2×2 determinant.
\[ \begin{vmatrix} 3 - \lambda & 2 \\ 2 & -3 - \lambda \end{vmatrix} = (3 - \lambda)(-3 - \lambda) - 4 = [-(9 - \lambda^2) - 4] = \lambda^2 - 13 \]

Hence the characteristic equation becomes:
\[ (-3 - \lambda)(\lambda^2 - 13) = 0 \]

Step 4: Find eigenvalues.
\[ \lambda = -3, \quad \lambda = \sqrt{13}, \quad \lambda = -\sqrt{13} \]

Step 5: Determine imaginary parts.
All the eigenvalues are real numbers.
Hence, their imaginary parts are 0, 0, 0.

Final Answer: 0, 0, 0