ghGiven that $A=\begin{bmatrix} 3&3 &0 \\ 3& 3& 0\\0 &0 &6 \end{bmatrix}$
We can write characteristic equation $|A-\lambda I|=0$
$\begin{vmatrix} 3-\lambda&3&0 \\ 3&3-\lambda &0 \\ 0& 0& 6-\lambda\end{vmatrix}=0$
$(3-\lambda)[(3-\lambda)(6-\lambda)-0]-3[3(6-\lambda)-0]+0=0$
$(3-\lambda)(3-\lambda)(6-\lambda)-9(6-\lambda)=0$
$(3-\lambda)^{2}(6-\lambda)-9(6-\lambda)=0$
$(6-\lambda)[(3-\lambda)^{2}-9]=0$
$(\lambda-6)[(\lambda-3)^{2}-9]=0$
$(\lambda-6)[\lambda^{2}+9-6\lambda-9]=0$
$(\lambda-6)[\lambda^{2}-6\lambda]=0$
$\lambda(\lambda-6)(\lambda-6)=0$
So, we got characteristic roots $\lambda=0,6,6$
Now,$f(x)=x(x-6)^{2}$
So$,g(x)=x(x-6)$
By Cayley Hamilton theorem:-Every square matrix satisfy its characteristic equation.
$f(x)=0$
$(or)$
$f(A)=0$
Lets verify $g(A)$ is zero or not$?$
$g(x)=x(x-6)=x^{2}-6x$
$g(A)=A^{2}-6A$
$g(A)=\begin{bmatrix} 18&18 &0 \\18 & 18& 0\\0 &0 &36 \end{bmatrix}-6\begin{bmatrix} 3&3 &0 \\ 3& 3& 0\\0 &0 &6 \end{bmatrix}$
$g(A)=\begin{bmatrix} 18&18 &0 \\18 & 18& 0\\0 &0 &36 \end{bmatrix}-\begin{bmatrix} 18&18 &0 \\ 18& 18& 0\\0 &0 &36 \end{bmatrix}$
$g(A)=0$
$A^{2}-6A=0$
also $h(x)=x$ $;A\neq0$
$p(x)=(x-6)$ $;A-6\neq0$
g(x) is the monic polynomial of lowest degree such that $g(x)=0$
So,$g(x)=x(x-6)$ is the minimal polynomial.