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 $= -24$ .

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.
