Question

Consider the 5 × 5 matrix

\(A = \left[ {\begin{array}{*{20}{c}} 1&2&3&4&5\\ 5&1&2&3&4\\ 4&5&1&2&3\\ 3&4&5&1&2\\ 2&3&4&5&1 \end{array}} \right]\).

It is given that A has only one real eigenvalue. Then the real eigenvalue of A is

1. -2.5
2. 0
3. 15
4. 25

Answer 1

Correct Answer - Option 3 : 15

\(A = \left[ {\begin{array}{*{20}{c}} 1&2&3&4&5\\ 5&1&2&3&4\\ 4&5&1&2&3\\ 3&4&5&1&2\\ 2&3&4&5&1 \end{array}} \right]\)

The Eigenvalues of a Matrix are given by:

|A – λ I| = 0

\(\left[ {\begin{array}{*{20}{c}} {1 - \lambda }\\ {\begin{array}{*{20}{c}} 5\\ {\begin{array}{*{20}{c}} 4\\ {\begin{array}{*{20}{c}} 3\\ 2 \end{array}} \end{array}} \end{array}} \end{array}\begin{array}{*{20}{c}} 2\\ {\begin{array}{*{20}{c}} {1 - \lambda }\\ {\begin{array}{*{20}{c}} 5\\ 4\\ 3 \end{array}} \end{array}} \end{array}\begin{array}{*{20}{c}} 3\\ {\begin{array}{*{20}{c}} 2\\ {\begin{array}{*{20}{c}} {1 - \lambda }\\ 5\\ 4 \end{array}} \end{array}} \end{array}\begin{array}{*{20}{c}} 4\\ {\begin{array}{*{20}{c}} 3\\ {\begin{array}{*{20}{c}} 2\\ {1 - \lambda }\\ 5 \end{array}} \end{array}} \end{array}\begin{array}{*{20}{c}} 5\\ {\begin{array}{*{20}{c}} 4\\ {\begin{array}{*{20}{c}} 3\\ 2\\ {1 - \lambda } \end{array}} \end{array}} \end{array}} \right] = 0\)

Apply Row Transformation

R₁ → R₁ + R₂ + R₃ + R₄ + R₅

\(\left[ {\begin{array}{*{20}{c}} {15 - \lambda }\\ {\begin{array}{*{20}{c}} 5\\ {\begin{array}{*{20}{c}} 4\\ {\begin{array}{*{20}{c}} 3\\ 2 \end{array}} \end{array}} \end{array}} \end{array}\begin{array}{*{20}{c}} {\;15 - \lambda }\\ {\begin{array}{*{20}{c}} {1 - \lambda }\\ {\begin{array}{*{20}{c}} 5\\ 4\\ 3 \end{array}} \end{array}} \end{array}\begin{array}{*{20}{c}} {\;15 - \lambda \;}\\ {\begin{array}{*{20}{c}} 2\\ {\begin{array}{*{20}{c}} {1 - \lambda }\\ 5\\ 4 \end{array}} \end{array}} \end{array}\begin{array}{*{20}{c}} {15 - \lambda \;}\\ {\begin{array}{*{20}{c}} 3\\ {\begin{array}{*{20}{c}} 2\\ {1 - \lambda }\\ 5 \end{array}} \end{array}} \end{array}\begin{array}{*{20}{c}} {15 - \lambda }\\ {\begin{array}{*{20}{c}} 4\\ {\begin{array}{*{20}{c}} 3\\ 2\\ {1 - \lambda } \end{array}} \end{array}} \end{array}} \right]\)

\(\left( {15 - \lambda } \right)\left( {\begin{array}{*{20}{c}} 1\\ {\begin{array}{*{20}{c}} 5\\ {\begin{array}{*{20}{c}} 4\\ {\begin{array}{*{20}{c}} 3\\ 2 \end{array}} \end{array}} \end{array}} \end{array}\begin{array}{*{20}{c}} 1\\ {\begin{array}{*{20}{c}} {\;1 - \lambda }\\ {\begin{array}{*{20}{c}} 5\\ 4\\ 3 \end{array}} \end{array}\;} \end{array}\begin{array}{*{20}{c}} 1\\ {\begin{array}{*{20}{c}} 2\\ {\begin{array}{*{20}{c}} {1 - \lambda }\\ 5\\ 4 \end{array}\;} \end{array}} \end{array}\begin{array}{*{20}{c}} 1\\ {\begin{array}{*{20}{c}} 3\\ {\begin{array}{*{20}{c}} {\;2}\\ {1 - \lambda }\\ 5 \end{array}} \end{array}} \end{array}\begin{array}{*{20}{c}} 1\\ {\begin{array}{*{20}{c}} 4\\ {\begin{array}{*{20}{c}} 3\\ 2\\ {\;1 - \lambda } \end{array}} \end{array}} \end{array}} \right)\)

It is given that the matrix A has only one real eigenvalue, i.e.

⇒ 15 - λ = 0

⇒ λ = 15

The other 4 eigenvalues are complex conjugates .