Characteristic equation is | A- λ I | =0

$ \left| \begin{array}{ccc} 2-\lambda & 2 & 1 \\ 1 & 3-\lambda & 1 \\ 1 & 2 & 2-\lambda \end{array}\right| $

∴ (2- λ) [(3-λ)(2-λ)-2]-2[(2-λ)+1]+1[2-(3-λ)]=0

∴$ (2- λ)[6-5λ+ λ^2-2]-2[1-λ]+ λ-1=0$

∴ $(2- λ)[ λ^2-5λ+4]+2λ-2+ λ-1=0$

∴ $2λ^2-10λ+8-λ^3+5λ^2-4λ+3λ-3=0$

∴$-λ^3+7λ^2-11λ+5=0$

∴ λ=5,1,1

∴ The eigen values λ are 5,1,1

Case 1 , when λ=5

| A- λ I | X=0

$ \left[ \begin{array}{ccc} -3 & 2 & 1 \\ 1 & -2 & 1 \\ 1 & 2 & -3 \end{array}\right] \left[ \begin{array}{ccc} X_1 \\ X_2 \\ X_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array}\right] $

∴ $-3 X_1+2 X_2+X_3 = 0$ and $- X_1-2 X_(2 )+X_3 = 0$

Using Cramer’s Rule :

$ \dfrac{X_1}{ \left| \begin{array}{cc} -2 & 1 \\ -2 & 1 \end{array}\right| } \;=\; \dfrac{X_2}{ \left| \begin{array}{cc} -3 & 1 \\ 1 & 1 \end{array}\right| } \;=\; \dfrac{X_3}{ \left| \begin{array}{cc} -3 & 2 \\ 1 & -2 \end{array}\right| } \\ \dfrac{X_1}{ 4} \;=\; \dfrac{-X_2}{ -4} \;=\; \dfrac{X_3}{ 4} $

∴ Eigen vector $X_1=[1 \; 1 \; 1 ]$ for eigen value 5

Case 2 , when λ=1

| A- λ I | X=0

$ \left[ \begin{array}{ccc} 1 & 2 & 1 \\ 1 & 2 & 1 \\ 1 & 2 & 1 \end{array}\right] \left[ \begin{array}{ccc} X_1 \\ X_2 \\ X_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array}\right] $

∴ $R_3- R_1$ and $R_2- R_1$ , we get

∴ $X_1+2 X_2+X_3 = 0$ ---------------(1)

∴ Here rank = 1 and number of unknown = 3

∴ 3 -1 = 2

∴ Let $X_1=1$ and $X_2=0$

∴ $X_3= -1$

∴ Eigen vector $X_2$=[1 ,0 ,-1 ]

Similarly , Let $X_1$=0 and $X_2$=1

∴ $X_3$=2

∴ Eigen vector $X_3$=[0 ,1 ,2 ]

∴ Eigen vector is [1 0 -1] and [0 1 2 ] for eigen value 1

Given f(A)= $A^3$+1

∴ f(λ)= $λ^3$+1

When λ=5 ,f(5)= $5^3$+1=126

When λ=1 ,f(1)= $1^3$+1=2

∴ Eigen values of $A^3$+I are 126 ,2,2 and

Eigen vectors are

$ X_1$=[1 1 1 ]

$X_2$=[1 ,0 ,-1 ]

$X_3$=[0 ,1 ,2 ]