Given A = [ 2 2 − 1 0 0 1 0 − 4 5 ] . {\displaystyle A=\left[{\begin{array}{ccc}2&2&-1\\0&0&1\\0&-4&5\end{array}}\right].} We solve d e t ( A − λ I ) = 0 {\displaystyle det(A-\lambda I)=0} to find λ . {\displaystyle \lambda .} So,
d e t ( [ 2 2 − 1 0 0 1 0 − 4 5 ] − λ [ 1 0 0 0 1 0 0 0 1 ] ) = d e t [ 2 − λ 2 − 1 0 − λ 1 0 − 4 5 − λ ] = 0 {\displaystyle det(\left[{\begin{array}{ccc}2&2&-1\\0&0&1\\0&-4&5\end{array}}\right]-\lambda \left[{\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}}\right])=det\left[{\begin{array}{ccc}2-\lambda &2&-1\\0&-\lambda &1\\0&-4&5-\lambda \end{array}}\right]=0}
We have
d e t [ 2 − λ 2 − 1 0 − λ 1 0 − 4 5 − λ ] = ( 2 − λ ) [ − λ ( 5 − λ ) − ( − 4 ) ] = ( 2 − λ ) [ λ 2 − 5 λ + 4 ] = ( 2 − λ ) ( λ − 1 ) ( λ − 4 ) = 0. {\displaystyle det\left[{\begin{array}{ccc}2-\lambda &2&-1\\0&-\lambda &1\\0&-4&5-\lambda \end{array}}\right]=(2-\lambda )[-\lambda (5-\lambda )-(-4)]=(2-\lambda )[\lambda ^{2}-5\lambda +4]=(2-\lambda )(\lambda -1)(\lambda -4)=0.}
Solving the above cubic equation, we get λ 1 = 1 , λ 2 = 2 , λ 3 = 4. {\displaystyle \color {blue}\lambda _{1}=1,\lambda _{2}=2,\lambda _{3}=4.}
![{\displaystyle A=\left[{\begin{array}{ccc}2&2&-1\\0&0&1\\0&-4&5\end{array}}\right].}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/038378d22093e9ad67aeedebe46e1a656d260aae)
We solve 
to find 
So,
![{\displaystyle det(\left[{\begin{array}{ccc}2&2&-1\\0&0&1\\0&-4&5\end{array}}\right]-\lambda \left[{\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}}\right])=det\left[{\begin{array}{ccc}2-\lambda &2&-1\\0&-\lambda &1\\0&-4&5-\lambda \end{array}}\right]=0}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/1b7c018c983b8d21eb58fc5dfe95b4fd972fb41d)
![{\displaystyle det\left[{\begin{array}{ccc}2-\lambda &2&-1\\0&-\lambda &1\\0&-4&5-\lambda \end{array}}\right]=(2-\lambda )[-\lambda (5-\lambda )-(-4)]=(2-\lambda )[\lambda ^{2}-5\lambda +4]=(2-\lambda )(\lambda -1)(\lambda -4)=0.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/b0b8b78079ac1ab17ec4173eea828b33ccf842e0)
