To find eigen values & eigen vectors $A^3 +I$ first we find eigen values & eigen vectors of A Characteristic eqation is |A- λI| =0

i.e. $$ \begin{vmatrix} 2-λ&2&1 \\ 1&3-λ&1 \\ 1&2&2-λ \end{vmatrix} = 0 $$

Expanding we get a cubic equation in λ as $λ^3 - 7λ^2 + 11λ - 5 =0$ i.e. λ = 1, 1, 5

To find eigen vectors corresponding to each eigen value we write matrix equation [A- λI]X =0 ,Where X is eigen vector (which is a column matrix)

Let, $X= \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} $, therefore matrix equation can be written as: $$ \begin{bmatrix} 2-λ&2&1 \\ 1&3-λ&1 \\ 1&2&2-λ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} =0………….(1) $$

(i) For λ= 1 ,equation (1) becomes:
$$ \begin{bmatrix} 1&2& 1 \\ 1&2& 1 \\ 1&2&1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} =0 $$

If for repeated eigen values we get all the rows same then we can cot apply cramer’s rule ,we expand only $R_1$ ,we get $x_1 +2x_2 + x_3 =0 $

Now we have a single equation containing three variables ,to get the solution we put two convenient values to the last two variables (here we put $x_2 =1$ & $x_3 =0$) and find the value of first one ,therefore $x_1 = -2$

Therefore for λ= 1 , $$ X= \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} $$

For repeated λ= 1 again we get the same equation i.e. $x_1 +2x_2 + x_3 =0$ ,now to get the solution we interchange the values of last two variables (here we put $x_2 =0$ & $x_3 =1$) and find the value of first one ,therefore $x_1 = -1$

Therefore for repeated λ= 1, $$ X= \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} $$

(ii) For λ= 5 ,equation (1) becomes: $$ \begin{bmatrix} -3& 2& 1 \\ 1&-2& 1 \\ 1&2&-3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} =0 $$

To find $x_1 ,x_2 , x_3$ expanding to different rows $R_1$ & $R_2$ from above eqn & applying cramer’s rule ,we get:

$-3x_1 +2x-2 +x_3 =0 $ and $ x_1 - 2x_2 + x_3 =0$

By cramer’s rule: $$ \frac{x_1}{ \begin{vmatrix} 2&1 \\ -2&1 \end{vmatrix}} = \frac{-x_2}{ \begin{vmatrix} -3&1 \\ 1&1 \end{vmatrix}} = \frac{x_3}{ \begin{vmatrix} -3&2 \\ 1&-2 \end{vmatrix}} $$ Or $$ \frac{x_1}{4} =\frac{-x_2}{(-4)}= \frac{x_3}{4} $$

Therefore for λ= 5 , $X= \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} $

Therefore eigen values of $A_3+I$ are $λ^3+ 1$ or 2 , 2, 126 & corresponding eigen vectors are $\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} , \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} , \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} $

NOTE : (1) Eigen values may be repeated or distinct but corresponding to each eigen value there must be different eigen vector (exceptional case is there eg. 7)

(2) If λ is the eigen value of a matrix A and corresponding eigen vector is X then any equation in terms of A will have same eigen vector i.e. X

(3) To find eigen values, characteristic eqn of order 3 can be expanded using the following formula

$λ^3 –(sum \ of \ diagonal \ elements \ of \ A)λ^2$

+$(sum \ of \ minors \ of \ diagonal \ elements \ of \ A)λ - |A| =0$

(4) Eigen values of triangular matrices are same as its diagonal elements.