Using row operations to compute the following 3x3 determinant 
Use row operations to compute the following determinant $\begin{bmatrix}3&3&-3\\3&4&-4\\2&-3&-5\end{bmatrix}$

I know how to easily compute the determinant using $i - j + k$ method... The problem is I have put the matrix in LTF (Lower Triangular Form) and then used the product of the diagonal. So I did:
$R_2 \leftarrow R_1 + (-1)R_2$
$R_3 \leftarrow 2R_1 + (-3)R_3$
$R_3 \leftarrow 15R_2 + R_3$
to get the matrix in LTF $\begin{bmatrix}3&3&-3\\0&-1&1\\0&0&23\end{bmatrix} \implies \det(A) = (3)(-1)(23) = -69$ I know that since I used these operations I changed the determinant, how exactly can I fix it back? I also know that the $\det(A)$ should be $= -24$ .
 A: Multiplying any row by $\lambda$ multiplies the determinant by $\lambda$.
Adding any multiple of a row to another row doesn't change the determinant.
So

$R_2 \leftarrow R_1 + (-1)R_2$

multiplies the determinant by $-1$ (because multiplying $R_2$ by $-1$ multiplies the determinant by $-1$ and adding $R_1$ to $R_2$ does nothing to the determinant).

$R_3 \leftarrow 2R_1 + (-3)R_3$

multiplies the determinant by $-3$. 

$R_3 \leftarrow 15R_2 + R_3$

multiplies the determinant by $1$ (i.e. does nothing).
Overall the determinant has been multiplied by a factor of $-1\times-3\times1=3$. So dividing the new determinant by $3$ will give the original determinant.
A: The final matrix should be
$$
\begin{bmatrix}
3 & 3 & -3 \\
0 & -1 & 1 \\
0 & 0 & 24
\end{bmatrix}
$$
However, you have multiplied the determinant by $-1$ with the first operation and by $-3$ with the second one, so you get
$$
\frac{3\cdot(-1)\cdot24}{(-1)\cdot(-3)}=-24
$$
I use a different method, reducing the pivots to $1$:
\begin{align}
\begin{bmatrix}
3 & 3 & -3 \\
3 & 4 & -4 \\
2 & -3 & -5
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1 & -1 \\
3 & 4 & -4 \\
2 & -3 & -5
\end{bmatrix} && R_1\gets\color{red}{\frac{1}{3}}R_1
\\[6px]
&\to
\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & -1 \\
2 & -3 & -5
\end{bmatrix} && R_2\gets R_2-R_1
\\[6px]
&\to
\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & -1 \\
0 & -5 & -3
\end{bmatrix} && R_3\gets R_3-2R_1
\\[6px]
&\to
\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & -1 \\
0 & 0 & -8
\end{bmatrix} && R_3\gets R_3+5R_2
\\[6px]
&\to
\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & -1 \\
0 & 0 & 1
\end{bmatrix} && R_3\gets \color{red}{-\frac{1}{8}}R_3
\end{align}
The red numbers tell that the determinant has been multiplied by
$$
-\frac{1}{24}
$$
so it is $-24$.
A: The problem is that the operations you did were not elementary row operations, but rather compound operations that involved multiplying the individual rows before performing a row operation.
When you are doing the operations, you can swap two rows (that changes the determinant by a factor of $-1$), and you can replace a row $R_i$ with $R_i+aR_j$ for some different $R_j$, but you cannot replace $R_i$ with $aR_i+bR_j$ for any $a\neq 1$, because that is the equivalent to multiplying $R_i$ by $a$ first, which multiplies the determinant by $a$.  
Of course, if you are careful to keep track of all the ways you are multiplying, you can keep track of what you're doing to the determinant, and then compensate at the end.
