Question For each of the following matrices A perform the following tasks: (i) Find eigenvalues of A. (ii) Determine an eigenbasis. 4 -3 (1) A 8 -6 1 -1 (2) A= 1 2 1 2 1 -1 1 0 :

(1)A=\left[\begin{array}{cc}4 & -3 \\8 & -6\end{array}\right](i) Finding eigen values of A.\begin{array}{l}{[A-\lambda I]=0} \\{\left[\begin{array}{cc}4 & -3 \\8 & -6\end{array}\right]-\left[\begin{array}{ll}\lambda & 0 \\0 & \lambda\end{array}\right]=0} \\{\left[\begin{array}{cc}4-\lambda & -3 \\8 & -6-\lambda\end{array}\right]=0 .} \\(4-\lambda)(-6-\lambda)-(-3) 8=0 \\-24-4 \lambda+6 \lambda+\lambda^{2}+24=0 \\\lambda 2+2 \lambda=0 \\\lambda(\lambda+2)=0 \Rightarrow \lambda=0,-2 .\end{array}\therefore eigen values are \lambda_{1} \& \lambda_{2} are 0,-2.(ii) cigen basis of \stackrel{\underline{=}}{=} :before eigen base we need to find eigen-Vectors of \lambda.eigen rector for \lambda=0.\begin{array}{l}{[A-\lambda I]=0} \\{\left[\begin{array}{ll}4 & -3 \\8 & -6\end{array}\right]\left[\begin{array}{l}u_{1} \\u_{2}\end{array}\right]=0 .}\end{array}we need to reduced in echlon formR_{1}: R_{1} \rightarrow \frac{R_{1}}{4} ; R_{2}: R_{2}-8 R_{1} \text {. }\begin{array}{c}{\left[\begin{array}{cc}1 & -\frac{3}{4} \\0 & 0\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]=\left[\begin{array}{l}0 \\0\end{array}\right]} \\x_{1}-\frac{3}{4} x_{2}=0 \\x_{1}=\frac{3}{4} x_{2}\end{array}let x_{2}=t, x_{1}=\frac{3}{4} t\text { 荃 }\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]=\left[\begin{array}{c}\frac{3}{4} t \\t\end{array}\right]=\left[\begin{array}{l}\frac{3}{4} \\1\end{array}\right] t\stackrel{1}{\text { eigen veetor for }} \lambda=0.\begin{array}{l}{[A-\lambda I][x]=0} \\{\left[\begin{array}{cc}6-4 & -3 \\8 & -6-(-2)\end{array}\right]\left[\begin{array}{l}u_{1} \\u_{2}\end{array}\right]=0}\end{array}R: \quad R_{1} \rightarrow \frac{R_{1}}{6} ; R_{2}: R_{2}-8 R_{1}\begin{array}{l}{\left[\begin{array}{cc}1 & -1 / 2 \\0 & 0\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]=0} \\{\left[\begin{array}{l}u_{1} \\u_{2}\end{array}\right]=\left[\begin{array}{c}1 / 2 \\1\end{array}\right] t .}\end{array}Now These eigen vectors are Arranged into a new matriu called eigen base.=\left[\begin{array}{cc}\frac{3}{4} & 1 / 2 \\1 & 1\end{array}\right](2)A=\left[\begin{array}{ccc}1 & -1 & 0 \\1 & 2 & 1 \\-2 & 1 & -1\end{array}\right](i) eigon values of : :\begin{array}{l}{[A-\lambda J]=0} \\{\left[\begin{array}{ccc}1-\lambda & -1 & 0 \\1 & 2-\lambda & 1 \\-2 & 1 & -1-\lambda\end{array}\right]=0} \\\Rightarrow(1-\lambda)[(2-\lambda)(-1-\lambda)-1]-(-1)[(-1-\lambda) 1-(-2)]+0=0 \\\Rightarrow(1-\lambda)\left[-2-2 \lambda+\lambda+\lambda^{2}-1\right]+1[-1-\lambda+2]=0 \\\Rightarrow-\lambda^{3}+2 \lambda^{2}+\lambda-2=0 \text {. } \\\end{array}we get \lambda_{1}=2, \lambda_{2}=1, \lambda_{3}=-1.\therefore eigen values are 2,1,-1.(ii) eigen basis of A:Befor that we need to caluclate eigen rectors for eigen values.for \lambda=2\left[\begin{array}{ccc}1-\lambda & -1 & 0 \\1 & 2-\lambda & 1 \\-2 & 1 & -\lambda-1\end{array}\right]=\left[\begin{array}{ccc}-1 & -1 & 0 \\1 & 0 & 1 \\-2 & 1 & -3\end{array}\right]\left[\begin{array}{l}u_{1} \\u_{2} \\u_{3}\end{array}\right]=0we need to convert it into row echlon form\begin{array}{l}R_{1}: R_{1} \longrightarrow R_{1} \cdot(-1) \\\text { [ multiply row } 1 \text { by -1] } \\R_{2}: R_{2} \rightarrow R_{2}-R_{1} \\\end{array}\begin{array}{l}{\left[\begin{array}{ccc}1 & 1 & 0 \\0 & -1 & 1 \\0 & 3 & -3\end{array}\right]} \\R_{2} \longrightarrow-R_{2} \quad\left[\text { multiply row } 2 b_{y}-1\right] \\R_{1} \longrightarrow R_{1}-R_{2} \\{\left[\begin{array}{ccc}1 & 0 & 1 \\0 & 1 & -1 \\0 & 3 & -3\end{array}\right]} \\R_{3}=R_{3}-3 R_{2} \\{\left[\begin{array}{ccc}1 & 0 & 1 \\0 & 1 & -1 \\0 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2} \\u_{3}\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right]} \\x_{1}+x_{3}=0 ; \quad x_{2}=x_{3} \\x_{3}=-x_{1} \\\end{array}iet, x_{3}=tfor \lambda=1\left[\begin{array}{l}u_{1} \\u_{2} \\u_{3}\end{array}\right]=\left[\begin{array}{c}-t \\t \\t\end{array}\right]=\left[\begin{array}{c}-1 \\1 \\1\end{array}\right] t\begin{array}{l}{\left[\begin{array}{ccc}1-1 & -1 & 0 \\1 & 2-1 & 1 \\-2 & 1 & -1-1\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2} \\x_{3}\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right]} \\{\left[\begin{array}{ccc}0 & -1 & 0 \\1 & 1 & 1 \\-2 & 1 & -2\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2} \\x_{3}\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right]}\end{array}R_{1}=R_{1} \leftrightarrow R_{2} \quad [Row 2 \& 1 are interchanging]R_{3}=R_{3}+2 R_{1}\begin{array}{l}{\left[\begin{array}{ccc}1 & 1 & 1 \\ 0 & -1 & 0 \\ 0 & 3 & 0\end{array}\right]\left[\begin{array}{l}u_{1} \\ u_{2} \\ u_{3}\end{array}\right]=[0]} \\ R_{2}=-R_{2} \quad\left[\text { multiply Row } 2 b_{y}-1\right] \\ R_{1}=R_{1}-R_{2} \\ {\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 3 & 0\end{array}\right]} \\ R_{3}=R_{3}-3 R_{2} \\ {\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{array}\right]\left[\begin{array}{l}u_{1} \\ u_{2} \\ u_{3}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]} \\ u_{2}=0 \quad x_{1}+u_{3}=0 \\ x_{1}=-x_{3} \\ {\left[\begin{array}{l}u_{1} \\ u_{2} \\ u_{3}\end{array}\right]=\left[\begin{array}{c}-t \\ 0 \\ t\end{array}\right]=\left[\begin{array}{c}-1 \\ 0 \\ 1\end{array}\right] t} \\ \lambda=-1:- \\ {\left[\begin{array}{ccc}1-(-1) & -1 & 0 \\ 1 & 2-(-1) & 1 \\ -2 & 1 & -(x-1)-1\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ u_{3}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]} \\ {\left[\begin{array}{ccc}2 & -1 & 0 \\ 1 & 3 & 1 \\ -2 & 1 & 0\end{array}\right]\left[\begin{array}{l}u_{1} \\ x_{2} \\ u_{3}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]} \\ R_{1}=R_{1} / 2 \\ R_{2}=R_{2}-R_{1} \\\end{array}\begin{array}{l}{\left[\begin{array}{ccc}1 & -1 / 2 & 0 \\0 & 7 / 2 & 1 \\-2 & 1 & 0\end{array}\right]} \\R_{3}=R_{3}+2 R_{1} \\R_{2}=2 R_{2} / 7 \\{\left[\begin{array}{ccc}1 & -1 / 2 & 0 \\0 & 1 & 2 / 7 \\0 & 0 & 0\end{array}\right]} \\R_{1}=R_{1}+\frac{R_{2}}{2} \\{\left[\begin{array}{lcc}1 & 0 & 1 / 7 \\0 & 1 & 2 / 7 \\0 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2} \\u_{3}\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right]} \\u_{1}+u_{3} / 7=0 \\\text { let, } x_{3}=t . \quad x_{2}+\frac{2 u_{3}}{7}=0 \\{\left[\begin{array}{l}x_{1} \\x_{2} \\x_{3}\end{array}\right]=\left[\begin{array}{c}-\frac{t}{7} \\-\frac{2 t}{7} \\t\end{array}\right]=\left[\begin{array}{c}-1 / 7 \\-2 / 7 \\1\end{array}\right] t} \\\end{array}(3) IN Case if you have any doubt feel free to comment Pls support my answer with a like ...