- Characteristic equation of A is $\mid A-\lambda I \mid = 0$
$\lambda^3-(3+5+3) \lambda^2 + [(15+15+9)-(1+1+1)]\lambda - \mid A \mid =0$
$\lambda^3-11\lambda^2+36\lambda-36=0$
$\lambda=6,3,2$ are eigen values of A
Now to find eigen vector $X\neq0$ such that $[A-\lambda I]=0$
$ \left[ {\begin{array}{cc}
3-\lambda & -1 & 1 \\
-1 & 5-\lambda & -1 \\
1 & -1 & 3-\lambda
\end{array} } \right]$ $ \left[ {\begin{array}{cc}
x \\
y \\
z
\end{array} } \right]$= $ \left[ {\begin{array}{cc}
0 \\
0 \\
0
\end{array} } \right]$ (1)
Put $\lambda=6$ in (1)
$ \left[ {\begin{array}{cc}
-3 & -1 & 1 \\
-1 & -1 & -1 \\
1 & -1 & 3
\end{array} } \right]$ $ \left[ {\begin{array}{cc}
x \\
y \\
z
\end{array} } \right]$= $ \left[ {\begin{array}{cc}
0 \\
0 \\
0
\end{array} } \right]$
Choose any two distinct rows. Let us choose 2nd and 3rd row, we get,
$-x-y-z=0$
$x-y-3z=0$
By Cramer's rule,
$\frac{x}{\mid {\begin{array}{cc}
-1 & 1 \\
-1 & 3 \\
\end{array} } \mid}$= $\frac{-y}{\mid {\begin{array}{cc}
-1 & 1 \\
1 & 3 \\
\end{array} } \mid}$= $\frac{z}{\mid {\begin{array}{cc}
-1 & -1 \\
1 & -1 \\
\end{array} } \mid}$
$\frac{x}{2}=\frac{-y}{4}=\frac{z}{2}$
$ X_1=$
$ \left[ {\begin{array}{cc}
2 \\
-4 \\
2
\end{array} } \right]$ or $ \left[ {\begin{array}{cc}
1 \\
-2 \\
1
\end{array} } \right] $
Put $\lambda=3$ in (1) we get,
$\left[ {\begin{array}{cc}
0 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 0
\end{array} } \right]$$ \left[ {\begin{array}{cc}
x \
y \
z
\end{array} } \right]$=$ \left[ {\begin{array}{cc}
0 \\
0 \\
0
\end{array} } \right]$
From 2nd and 3rd row as they are distinct
$-x+2y-z=0$
$x-y+0=0$
By Cramer's rule,
$\frac{x}{\mid {\begin{array}{cc}
2 & -1 \\
-1 & 0 \\
\end{array} } \mid}$=$ \frac{x}{\mid {\begin{array}{cc}
-1 & 1 \\
1 & 0 \\
\end{array} } \mid}$= $\frac{x}{\mid {\begin{array}{cc}
-1 & 2 \\
1 & -1 \\
\end{array} } \mid}$
$\frac{x}{-1}=\frac{-y}{1}=\frac{z}{-1}$
$ X_2=$
$ \left[ {\begin{array}{cc}
-1 \\
-1 \\
-1
\end{array} } \right]$ or $ \left[ {\begin{array}{cc}
1 \\
1 \\
1
\end{array} } \right]$
Put $\lambda=2$ in (1), we get
$ \left[ {\begin{array}{cc}
1 & -1 & 1\\
-1 & 3 & -1 \\
1 & -1 & 1
\end{array} } \right] $ $ \left[ {\begin{array}{cc}
x\\
y \\
z
\end{array} } \right] $= $ \left[ {\begin{array}{cc}
0\\
0 \\
0
\end{array} } \right]$
$-x+3y-z=0$
$x-y+z=0$
$\frac{x}{\mid {\begin{array}{cc}
3 & -1 \\
-1 & 1\\
\end{array} } \mid}$ = $\frac{-y}{\mid {\begin{array}{cc}
-1 & 1 \\
1 & 1 \\
\end{array} } \mid}$= $\frac{z}{\mid {\begin{array}{cc}
-1 & 3 \\
1 & -1 \\
\end{array} } \mid}$
$\frac{x}{2}=\frac{-y}{0}=\frac{z}{-2} $
$ X_3=$
$ \left[ {\begin{array}{cc}
2\\
0 \\
-2
\end{array} } \right]$ or$ \left[ {\begin{array}{cc}
1\\
0 \\
-1
\end{array} } \right]$
Therefore,$\lambda=6,3,2 $ are eigen values of A and corresponding eigen vectors are
$ X_1=$
$ \left[ {\begin{array}{cc}
1\\
-2 \\
1
\end{array} } \right]$, $ X_2=$
$ \left[ {\begin{array}{cc}
1\\
1\\
1
\end{array} } \right]$ , $ X_3=$
$ \left[ {\begin{array}{cc}
2\\
0 \\
-2
\end{array} } \right]$
- Characteristic equation of A is $\mid A-\lambda I \mid = 0$
$\lambda^3-(6+3+3) \lambda^2 + [(18+18+9)-(4+4+1)]\lambda - \mid A \mid =0$
$\lambda^3-12\lambda^2+36\lambda-2=0$
$\lambda=8,2,2$ are eigen values of A
Now to find eigen vector $X\neq0$ such that $[A-\lambda I]=0$
$ \left[ {\begin{array}{cc}
6-\lambda & -2 & 2 \\
-2 & 3-\lambda & -1 \\
2 & -1 & 3-\lambda
\end{array} } \right]$ $ \left[ {\begin{array}{cc}
x \\
y \\
z
\end{array} } \right]$= $ \left[ {\begin{array}{cc}
0 \\
0 \\
0
\end{array} } \right]$ (1)
Put $\lambda=8$ in (1)
$ \left[ {\begin{array}{cc}
-2 & -2 & 2 \\
-2 & -5 & -1 \\
2 & -1 & -5
\end{array} } \right]$ $ \left[ {\begin{array}{cc}
x \\
y \\
z
\end{array} } \right]$= $ \left[ {\begin{array}{cc}
0 \\
0 \\
0
\end{array} } \right]$
Choose any two distinct rows. Let us choose 2nd and 3rd row, we get,
$-2x-5y-z=0$
$2x-y-5z=0$
By Cramer's rule,
$\frac{x}{\mid {\begin{array}{cc}
-5 & -1 \\
-1 & -5 \\
\end{array} } \mid}$= $\frac{-y}{\mid {\begin{array}{cc}
-2 & -1 \\
2 & -5 \\
\end{array} } \mid}$= $\frac{z}{\mid {\begin{array}{cc}
2 & -5 \\
2 & -1 \\
\end{array} } \mid}$
$\frac{x}{24}=\frac{-y}{12}=\frac{z}{12}$
$ X_1=$
$ \left[ {\begin{array}{cc}
24 \\
-12 \\
12
\end{array} } \right]$ or $ \left[ {\begin{array}{cc}
2 \\
-1 \\
1
\end{array} } \right] $
Put $\lambda=2$ in (1) we get,
$\left[ {\begin{array}{cc}
4 & -2 & 2 \\
-2 & 1 & -1 \\
2 & -1 & 1
\end{array} } \right]$$ \left[ {\begin{array}{cc}
x \
y \
z
\end{array} } \right]$=$ \left[ {\begin{array}{cc}
0 \\
0 \\
0
\end{array} } \right]$
Here, all rows are the same if we reduce it. Therefore, we will get only one equation.
$2x-y+z=0$
$x=y-2x$
Put $x=0$ and $x=1$
$y=1$ $z=1$ and $y=0$ and $z=-2$
$ X_2=$
$ \left[ {\begin{array}{cc}
0 \\
1 \\
1
\end{array} } \right]$ and $ X_3=$
$ \left[ {\begin{array}{cc}
1\\
0 \\
-2
\end{array} } \right]$
Therefore, $\lambda=8,2,2$ are eigen values of A and corresponding eigen vectors are $X_1$= $ \left[ {\begin{array}{cc}
2 \\
-1 \\
1
\end{array} } \right] $, $ X_2=$
$ \left[ {\begin{array}{cc}
0 \\
1 \\
1
\end{array} } \right]$ and $ X_3=$
$ \left[ {\begin{array}{cc}
1\\
0 \\
-2
\end{array} } \right]$