Correct Answer - Option 3 : 1, 1, 5
Concept:
Form the properties of Eigen values
i) Sum of eigen values = Trace of the matrix
ii) Product of Eigen values = Determinant of the matrix
Calculations:
Given:
\(\;A = \left[ {\begin{array}{*{20}{c}} 2&2&1\\ 1&3&1\\ 1&2&2 \end{array}} \right]\)
det (A) = {2 × (6 – 2)} – {2 × (2 – 1)} + {1 × (2 – 3)}
det (A) = (8 – 2 – 1)
det (A) = 5
Now as we know, Product of Eigen values = Determinant of the matrix
Only option 3 satisfy the given condition
1 × 1 × 5 = 5
Hence options 3 is correct.
\(\;A = \left[ {\begin{array}{*{20}{c}} 2&2&1\\ 1&3&1\\ 1&2&2 \end{array}} \right]\)
\(\left| {\begin{array}{*{20}{c}} 2-λ&2&1\\ 1&3-λ&1\\ 1&2&2-λ \end{array}} \right|=0\)
(2 - λ)[(3 - λ)(2 - λ) - 2] - 2[1(2 - λ) - 1] + 1[2 - 1(3 - λ)] = 0
-λ3 + 7λ2 - 11λ + 5 = 0
λ3 - 7λ2 + 11λ - 5 = 0
(λ - 1)(λ2 - 6λ + 5) = 0
(λ - 1)(λ - 1)(λ - 5) = 0
∴ λ = 1,1,5
