Solution:

$A=\begin{bmatrix} 4 & 1 & -1 \\ 2 & 5 & -2 \\ 1 & 1 & 2 \end{bmatrix}$

|A|=45

C.E.$|A-\lambda I|=0$

$A= \left| \begin{matrix} 4-\lambda & 1 & -1 \\ 2 & 5-\lambda & -2 \\ 1 & 1 & 2-\lambda \end{matrix} \right| =0$

$\lambda^3-11\lambda^2+(20+10+8-2+1+2)\lambda-45=0$

$\lambda^3-11\lambda^2+39\lambda-45=0$

$\lambda =3,3,5$

To find eigen vector

$[A-\lambda I]X =0$

$\begin{bmatrix} 4-\lambda & 1 & -1 \\ 2 & 5-\lambda & -2 \\ 1 & 1 & 2-\lambda \end{bmatrix}\begin{bmatrix} x_1\\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$------(1)

put $\lambda =3 $ in (1)

$\begin{bmatrix} 1& 1 & -1 \\ 2 & 2 & -2 \\ 1 & 1 & -1 \end{bmatrix}\begin{bmatrix} x_1\\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$

$R2-R1 \\ R3-\frac{R_2}{2}$ $\begin{bmatrix} 1& 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x_1\\ x_2 \\ x_3 \end{bmatrix} =\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$

$x_1+x_2-x_3=0$-------(2)

from (2)

$x_1=-x_2+x_3$

Put $x_2=-s \\ x_3=t$

$x_1=-(-s)+t$

$x_1=s+t$

from equation (2)

$x_2=-x1+x_3$

$x_2=-(s+t)+t$

$x_2=-s-t+t$

$x_2=-s$

From equation (2)

$x_3 = x_1+x_2$

$x_3= s+t-s$

$x_3=t$

$X_1 = \begin{bmatrix} s+t\\-s+0 \\ 0+t \end{bmatrix}$

$X_1 = \begin{bmatrix} s\\-s \\ 0 \end{bmatrix}$ +$\begin{bmatrix} t\\0 \\ t \end{bmatrix}$

$X_1 = s\begin{bmatrix} 1\\-1 \\ 0 \end{bmatrix} X_2=\begin{bmatrix} 1\\0 \\ 1\end{bmatrix}$

Put $\lambda=5$ in (1)

$\begin{bmatrix} 1& 1 & -1 \\ 2 & 0 & -2\\ 1 & 1 & -3\end{bmatrix}\begin{bmatrix} x_1\\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$

$-x_1+x_2-x_3=0$

$2x_1+0x_2-2x_3=0$

$x_1+x_2+x_3=0$

$\frac {x_1}{\begin{bmatrix} 1& -1 \\ 0 & -2 \end{bmatrix}}$ = $\frac {-x_2}{\begin{bmatrix} -1& -1 \\ 2 & -2 \end{bmatrix}}$ =$\frac {x_3}{\begin{bmatrix} -1& 1 \\ 2& 0 \end{bmatrix}}$

$\frac{x_1}{-2}= \frac{-x_2}{4}= \frac{x_3}{-2}$

$X_3 =\begin{bmatrix} -2\\-4 \\ -2 \end{bmatrix}$

$\therefore X_3 =\begin{bmatrix} 1\\2 \\ 1 \end{bmatrix}$

$M= [X_1, X_2, X_3]$

M=$\begin{bmatrix} 1 & 1 & 1 \\ -1& 0 & 2 \\ 0& 1 & 1 \end{bmatrix}$

$M^{-1}AM=D$

$A=\begin{bmatrix} 4 & 1 & -1 \\ 2 & 5 & -2 \\ 1 & 1 & 2 \end{bmatrix}$ will be diagonalised to the diagonal matrix, D=$\begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix}$ by the transformation matrix M=$\begin{bmatrix} 1 & 1 & 1 \\ -1 & 0& 2 \\ 0 & 1 & 1 \end{bmatrix}$