We first note that y ( t ) {\displaystyle y(t)} can be written as y ( t ) = [ i 1 ] e ( 2 + i ) t . {\displaystyle {\textbf {y}}(t)=\left[{\begin{array}{c}i\\1\end{array}}\right]e^{(2+i)t}.} Using the hint, we have v = [ i 1 ] {\displaystyle v=\left[{\begin{array}{c}i\\1\end{array}}\right]} and λ = 2 + i . {\displaystyle \lambda =2+i.} So, they have to satisfy ( A − λ I ) v = 0. {\displaystyle (A-\lambda I)v=0.}
Let A = [ a b c d ] . {\displaystyle A=\left[{\begin{array}{cc}a&b\\c&d\end{array}}\right].} We then get
0 = ( A − λ I ) v = ( [ a b c d ] − λ [ 1 0 0 1 ] ) [ i 1 ] = [ a − ( 2 + i ) b c d − ( 2 + i ) ] [ i 1 ] = [ ( 1 + b ) + ( a − 2 ) i ( d − 2 ) + ( c − 1 ) i ] . {\displaystyle 0=(A-\lambda I)v={\Big (}\left[{\begin{array}{cc}a&b\\c&d\end{array}}\right]-\lambda \left[{\begin{array}{cc}1&0\\0&1\end{array}}\right]{\Big )}\left[{\begin{array}{c}i\\1\end{array}}\right]=\left[{\begin{array}{cc}a-(2+i)&b\\c&d-(2+i)\end{array}}\right]\left[{\begin{array}{c}i\\1\end{array}}\right]=\left[{\begin{array}{c}(1+b)+(a-2)i\\(d-2)+(c-1)i\end{array}}\right].}
Therefore, ( 1 + b ) + ( a − 2 ) i = 0 {\displaystyle (1+b)+(a-2)i=0} and ( d − 2 ) + ( c − 1 ) i = 0. {\displaystyle (d-2)+(c-1)i=0.} This implies that a = 2 , b = − 1 , c = 1 {\displaystyle a=2,b=-1,c=1} and d = 2. {\displaystyle d=2.} We finally have
A = [ 2 − 1 1 2 ] . {\displaystyle \color {blue}A=\left[{\begin{array}{cc}2&-1\\1&2\end{array}}\right].}
![{\displaystyle {\textbf {y}}(t)=\left[{\begin{array}{c}i\\1\end{array}}\right]e^{(2+i)t}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/e6e698f044e8797e15ebc21915ce217575b7622d)
![{\displaystyle v=\left[{\begin{array}{c}i\\1\end{array}}\right]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/72327a517c185138b6b458f0790cd4da68fa40b6)
![{\displaystyle A=\left[{\begin{array}{cc}a&b\\c&d\end{array}}\right].}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/0aa9c3c59abe47114df490e557d24f3c9e6dfce1)
![{\displaystyle 0=(A-\lambda I)v={\Big (}\left[{\begin{array}{cc}a&b\\c&d\end{array}}\right]-\lambda \left[{\begin{array}{cc}1&0\\0&1\end{array}}\right]{\Big )}\left[{\begin{array}{c}i\\1\end{array}}\right]=\left[{\begin{array}{cc}a-(2+i)&b\\c&d-(2+i)\end{array}}\right]\left[{\begin{array}{c}i\\1\end{array}}\right]=\left[{\begin{array}{c}(1+b)+(a-2)i\\(d-2)+(c-1)i\end{array}}\right].}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/ce0eef50ea45053b0008557dc3dea7cf00d47486)
![{\displaystyle \color {blue}A=\left[{\begin{array}{cc}2&-1\\1&2\end{array}}\right].}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/37ba49f529de81661f44f7577fc6bcc722cab6bd)