\[\mathbf{B} = \left[\begin{array} {rrr} 2 & -1 \\ 1 & 1 \end{array}\right] \]
First, To find the eigen value, we need to construct the characteristic polynomial using below equation.
\[det(B−λI)=0\]
Substituting B and λI
\[det(\left[\begin{array} {rrr} 2 & -1 \\ 1 & 1 \end{array}\right] - \left[\begin{array} {rrr} λ & 0 \\ 0 & λ \end{array}\right])=0 \]
After subtracting λ
\[det(\left[\begin{array} {rrr} 2-λ & -1 \\ 1 & 1- λ \end{array}\right])=0 \]
After determining the determinent
\[(2-λ)(1-λ) + 1 = 0 \] This becomes
\[(2-λ)(1-λ) + 1 = 0 \]
After reducing, the eqn becomes a quatric eqn.
\[λ^2 -3λ +1 = 0 \]
Not finding the factors, trying to solve with quatric formulae. that is
\[ soln = {-b ±\sqrt({b^2 - 4ac})}/2a \] where b = -3 , a = 1 , c= 1
\[λ = (3 ± \sqrt(5))/2 \]
So eigen values comes as
\[ (5.53/2 , 1.23/2) \]