Prove a matrix over $\Bbb R$ is triangularizable

$$A=\begin{bmatrix}3&0&-2\\-2&0&1\\2&1&0\end{bmatrix}$$

Proof this matrix is triangularizable and determine the triangularisation.

Using the usual method I did

$$det(A-\lambda I)=0=\begin{bmatrix}3-\lambda&0&-2\\-2&-\lambda&1\\2&1&-\lambda\end{bmatrix}$$ I got $$-(\lambda-1)^3, \lambda=1$$

So it is triangularisable.

As my calculations went on, things didnt work out...

So from the above we have an eigenvector $\begin{bmatrix}1\\-1\\1\end{bmatrix}$ and a basis $\begin{bmatrix}1&0&0\\-1&1&0\\1&0&1\end{bmatrix}$

So $M_{B^*}=SM_BS^{-1}=\begin{bmatrix}1&0&-2\\0&0&-1\\0&1&2\end{bmatrix}$ where $S^{-1}$ is the basis we just determined.

Now we still don't have an upper triangular matrix. I was taught to do this whole thing again with a block matrix, in this case, $A_2=\begin{bmatrix}0&-1\\1&2\end{bmatrix}$

But then in the end I still didn't get an upper triangular matrix like i should...Where went wrong? I did everything with a calculator so the calculations should be correct.

EDIT: the second part is quite important i think thats where i screwd up. better show the calculations:

So same thing $det(A_2)=0$ I get a $\lambda=1$ and a eigenvector $\begin{bmatrix}-1\\1\end{bmatrix}$. Now switch this with the second vecotr in the basis vector we have $\begin{bmatrix}1&0&0\\-1&-1&0\\1&1&1\end{bmatrix}$ And from this we can determine the new basis vector $\begin{bmatrix}0\\-1\\1\end{bmatrix}=x_1(b_1)+x_2(e_2)+x_3(e_3)$ where $b_1$ is the eigenvector from the fist part, and $e_2$ and $e_3$ are standard basis vectors. And $x_1=0,x_2=-1,x_3=1$. That will be the second vector of the new basis vector. Now we have our $S^{-1}=S=\begin{bmatrix}1&0&0\\0&-1&0\\0&1&1\end{bmatrix}$

And again $M_{B^*}=SM_BS^{-1}$, which didn't give me the triangular matrix that i was looking for

you have not mentioned Jordan Form... When they give you all integer entries and eigenvalues, this is a good method. Your characteristic polynomial is $(x-1)^3.$ This becomes $(A-I)^3 = 0$ in matrices.

Next to check is the minimal polynomial. Indeed, $A-I \neq 0$ as well as $(A-I)^2 \neq 0.$ The minimal polynomial is also $(x-1)^3$ That is, there is just one genuine eigenvector and just one Jordan block, size 3 by 3.

I am going to construct a matrix $P$ such that $P^{-1} A P = J$ is in Jordan form. I start with a column vector $w$ with $(A-I)^2 w \neq 0.$ This is my choice, once I write out $(A-I)^2$ to see what is allowed. I took $w = (0,0,1)^T.$ This $w$ is going to be the right hand column of $P.$ The $T$ means transpose, I wrote as a transpose of a row.

The middle column is $v = (A-I)w.$ This came out $v = (-2,1,-1)^T$

The left hand column is $u = (A-I)v = (A-I)^2 w.$ Note how $(A-I) u = 0,$ it is the only genuine eigenvector direction.

$$ P = \left( \begin{array}{rrr} -2&-2&0 \\ 2&1&0 \\ -2&-1&1 \\ \end{array} \right) $$

The determinant of $P$ is $2.$ That is, $P^{-1}$ is $\frac{1}{2}$ times the adjoint of $A,$ the adjoint being all integers.