- Characteristic equation of A is $\mid A-\lambda I \mid=0$
$\lambda^3-(12)\lambda^2+36\lambda-32=0$
$\lambda=8,2,2.$
$(\lambda-8)(\lambda-2)(\lambda-2)=0$
Consider $(\lambda-8)(\lambda)=\lambda^2-0\lambda+16$
Now to check that $\lambda^2-10A+16I$
Therefore, consider $A^2-10A+16I$
$= \left[ {\begin{array}{cc}
44 & -20 & 20\\
20 & 14 & -10 \\
20 & -10 & 14\\
\end{array} } \right] $-$10\left[ {\begin{array}{cc}
6 & -2 & 2\\
-2 & 3 & -1 \\
2 & -1 & 3\\
\end{array} } \right] $+$16 \left[ {\begin{array}{cc}
1 & 0 & 0\\
0 & 1 & 0 \\
0 & 0 & 1\\
\end{array} } \right] $
$= \left[ {\begin{array}{cc}
44 & -20 & 20\\
20 & 14 & -10 \\
20 & -10 & 14\\
\end{array} } \right] $-$10\left[ {\begin{array}{cc}
60 & -20 & 20\\
-20 & 30 & -10 \\
20 & -10 & 30\\
\end{array} } \right] $+$16 \left[ {\begin{array}{cc}
16 & 0 & 0\\
0 & 16 & 0 \\
0 & 0 & 16\\
\end{array} } \right] $
=$16 \left[ {\begin{array}{cc}
0 & 0 & 0\\
0 & 0 & 0 \\
0 & 0 & 0\\
\end{array} } \right] $
$=\lambda^2-10\lambda+16$ annihilates A
Minimal polynomial is $\lambda^2-10\lambda+16$
A is derogatory.
- Characteristic equation of A is $\mid A-\lambda I \mid=0$
$\lambda^3-(6)\lambda^2+11\lambda-6=0$
$\lambda=1,2,3$
We have $(\lambda-1)(\lambda-2)(\lambda-3)=0$
All eigen values of A are distinct.
Minimal polynomial of $A=(\lambda-1)(\lambda-2)(\lambda-3)$=$\lambda^3-6\lambda^2+11\lambda-6$= characteristic polynomial.
A is non derogatory.