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. 
 A: Why divide, when simple addition and subtraction will do? Subtract twice the bottom row from the middle and top rows and you are left with
$$\begin{vmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 1 & 1 & 1\end{vmatrix} = 1-0+0=1$$
by evaluating by minors.
Even easier is if you take that matrix and add row 2 to row 3 and subtract row 1 to get
$$\begin{vmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & 0\end{vmatrix} = 1-0+0=1$$
or take this and switch the last two rows to get
$$-\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{vmatrix} = -(-1)=1$$
Your last step doesn't do what you said though, if you take twice the third row from the second you get
$$3\cdot\begin{vmatrix} 1 & 2/3 & 2/3 \\ 0 & 0 & -1 \\ 0 & 1/3 & 1/3\end{vmatrix}=3\cdot(-1)\cdot\begin{vmatrix} 1 & 2/3 & 2/3 \\ 0 & 1/3 & 1/3 \\ 0 & 0 & -1\end{vmatrix}=3/3=1$$
which is what you expect, since you divided the original row by $3$.
A: I find it a bit easier to avoid fractions, therefore I would avoid dividing by 3 first.  Instead, let's take 2R3-R2=R2 to obtain:
$\begin{bmatrix}3 & 2 & 2\\0 & 0 & 1\\1 & 1 & 1\end{bmatrix}$
Now, take R1-3R3=R3 to obtain
$\begin{bmatrix}3 & 2 & 2\\0 & 0 & 1\\0 & -1 & -1\end{bmatrix}$
Because we took 3R3, this will change the value of the determinant.  We need to compensate by multiplying by 1/3. Leaving us with:
1/3$\begin{bmatrix}3 & 2 & 2\\0 & 0 & 1\\0 & -1 & -1\end{bmatrix}$
Now, interchange R2 and R3 to obtain
1/3$\begin{bmatrix}3 & 2 & 2\\0 & -1 & -1\\0 & 0 & 1\end{bmatrix}$
This will change the value of the determinant.  We need to add a negative sign to compensate for the interchange.  Leaving us with:
-1/3$\begin{bmatrix}3 & 2 & 2\\0 & -1 & -1\\0 & 0 & 1\end{bmatrix}$
Now, multiply the diagonal and then multiply by -1/3 leaving you with 1.
A: 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
$$
3\cdot\frac{2}{3}\cdot\frac{1}{2}=1
$$
