<s>[INST] Finding the determinant using row operations. So I have to find the determinant of $\begin{bmatrix}3&2&2\\2&2&1\\1&1&1\end{bmatrix}$ using row operations. From what I've learned, the row operations that change the determinate are things like swaping rows makes the determinant negative and dividing a row by a value means you have to multiply it by that value. Once you have an upper triangular matrix then you just multiply the diagonal and you should have the determinant. So how come I can't get the right answer? Here's what I did. I start by dividing the first row by 3. $3\begin{bmatrix}1&2/3&2/3\\2&2&1\\1&1&1\end{bmatrix}$. Then I subtract the second row by 2 times the first row $3\begin{bmatrix}1&2/3&2/3\\0&2/3&-1/3\\1&1&1\end{bmatrix}$. Then I subtract the third and first row. $3\begin{bmatrix}1&2/3&2/3\\0&2/3&-1/3\\0&1/3&1/3\end{bmatrix}$. Then I multiply the third row by 2 and subtract it from the second row. $3\begin{bmatrix}1&2/3&2/3\\0&2/3&-1/3\\0&0&1\end{bmatrix}$. So now I have an upper triangular matrix so now I just do $(3)(1)(2/3)(1)$ and I get 2. The answer to this question however is not 2 but 1. So where did I go wrong? If the things I read were true then I should've got the right answer but I didn't which tells me that there's another row operation rule that no one told me about. 
 [/INST] Your error is when you multiply the third row by $2$ and subtract the second row; this introduces a factor $2$ that you have to remove.
Use a more systematic way:
\begin{align}
\begin{bmatrix}
3&2&2\\
2&2&1\\
1&1&1
\end{bmatrix}
&\to
\begin{bmatrix}
1&2/3&2/3\\
2&2&1\\
1&1&1
\end{bmatrix}
&&R_1\gets \tfrac{1}{3}R_1 &&\color{red}{3}
\\[4px]&\to
\begin{bmatrix}
1&2/3&2/3\\
0&2/3&-1/3\\
0&1/3&1/3
\end{bmatrix}
&& \begin{aligned}R_2&\gets R_2-2R_1\\R_3&\gets R_3-R_1\end{aligned}
\\[4px]&\to
\begin{bmatrix}
1&2/3&2/3\\
0&1&-1/2\\
0&1/3&1/3
\end{bmatrix}
&& R_2\gets\tfrac{3}{2}R_2 && \color{red}{\tfrac{2}{3}}
\\[4px]&\to
\begin{bmatrix}
1&2/3&2/3\\
0&1&-1/2\\
0&0&1/2
\end{bmatrix}
&&R_3\gets R_3-\tfrac{1}{3}R_2
\\[4px]&\to
\begin{bmatrix}
1&2/3&2/3\\
0&1&-1/2\\
0&0&1
\end{bmatrix}
&&R_3\gets 2R_3 &&\color{red}{\tfrac{1}{2}}
\end{align}
The determinant is
$$
3\cdot\frac{2}{3}\cdot\frac{1}{2}=1
$$
</s>