Find the eigenvalue for the matrix

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

Concept:

• Eigenvalues are a special set of scalars associated with a linear system of equations (i.e., a matrix equation) that are sometimes also known as characteristic roots, characteristic values.
• For a square matrix  A, an Eigenvector(X) and Eigenvalue(λ) make this equation true i.e., AX = λX.
• For eigen values, |A - λ × I| = 0

​Calculation:

Given:

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

|A - λI| = 0

⇒ \(\begin{vmatrix} 4~-~λ&1&3\\ 1&3~-~λ&1\\ 2&0&5~-~λ \end{vmatrix}\) = 0

⇒ (4 - λ)[(3 - λ)(5 - λ) -0] - 1[5 - λ - 2] + 3[0 - 2(3 - λ)] = 0

⇒ (4 - λ)(3 - λ)(5 -λ) - 3 + λ - 18 + 6λ = 0

⇒ \((60~+~5λ^2~-~35λ~-~12λ~-~λ^3~+~7λ^2)~+~7λ~-~21~=~0\)

⇒ \(λ^3~-~12λ^2~+~40λ~-~39~=~0\)

By hit and trial method, At λ = 3, The above equation becomes zero

So, (λ - 3) is the factor of the above equation.

∴ (λ - 3) (λ² - 9λ + 13) = 0 

⇒ λ = 3, \({9 \pm \sqrt{29} \over 2}\) = 7.19, 1.81

7.1 is only in the option so (3) is correct