<s>[INST] determinant calculation This question is in my assignment. We are not allowed to use any symbol to represent any elementary row and column operations used in the solution. We must solve it step-by-step. Please help me to check my solution word by word including my spelling and grammar.
Question: 
Given that
$$\begin{vmatrix}a& b& c\\ d& e& f\\ g& h& i\end{vmatrix}=2$$
find
$$\begin{vmatrix}3c-6a& 3b& a\\ 3i-9c+18a-6g& 3h-9b& g-3a\\ 3f-6d& 3e& d\end{vmatrix}.$$
Solution:
We interchange the second and third rows of the matrix $\begin{pmatrix}a& b& c\\ d& e& f\\ g& h& i\end{pmatrix}$ to get the matrix $\begin{pmatrix}a& b& c\\ g& h& i\\ d& e& f\end{pmatrix}$ and we have
$$\begin{vmatrix}a& b& c\\ g& h& i\\ d& e& f\end{vmatrix}=-\begin{vmatrix}a& b& c\\ d& e& f\\ g& h& i\end{vmatrix}=-2.$$
We interchange the first and third columns of the matrix $\begin{pmatrix}a& b& c\\ g& h& i\\ d& e& f\end{pmatrix}$ to get the matrix $\begin{pmatrix}c& b& a\\ i& h& g\\ f& e& d\end{pmatrix}$ and we have
$$\begin{vmatrix}c& b& a\\ i& h& g\\ f& e& d\end{vmatrix}=-\begin{vmatrix}a& b& c\\ g& h& i\\ d& e& f\end{vmatrix}=-(-2)=2.$$
We multiply the second column of the matrix $\begin{pmatrix}c& b& a\\ i& h& g\\ f& e& d\end{pmatrix}$ by 3 to get the matrix $\begin{pmatrix}c& 3b& a\\ i& 3h& g\\ f& 3e& d\end{pmatrix}$ and we have
$$\begin{vmatrix}c& 3b& a\\ i& 3h& g\\ f& 3e& d\end{vmatrix}=3\begin{vmatrix}c& b& a\\ i& h& g\\ f& e& d\end{vmatrix}=(3)(2)=6.$$
We add $(-2)$ times the third column of the matrix $\begin{pmatrix}c& 3b& a\\ i& 3h& g\\ f& 3e& d\end{pmatrix}$ to its first column to get the matrix $\begin{pmatrix}c-2a& 3b& a\\ i-2g& 3h& g\\ f-2d& 3e& d\end{pmatrix}$ and we have
$$\begin{vmatrix}c-2a& 3b& a\\ i-2g& 3h& g\\ f-2d& 3e& d\end{vmatrix}=\begin{vmatrix}c& 3b& a\\ i& 3h& g\\ f& 3e& d\end{vmatrix}=6.$$
We add $(-3)$ times the first row of the matrix $\begin{pmatrix}c-2a& 3b& a\\ i-2g& 3h& g\\ f-2d& 3e& d\end{pmatrix}$ to its second row to get the matrix $\begin{pmatrix}c-2a& 3b& a\\ i-3c+6a-2g& 3h-9b& g-3a\\ f-2d& 3e& d\end{pmatrix}$ and we have
$$\begin{vmatrix}c-2a& 3b& a\\ i-3c+6a-2g& 3h-9b& g-3a\\ f-2d& 3e& d\end{vmatrix}=\begin{vmatrix}c-2a& 3b& a\\ i-2g& 3h& g\\ f-2d& 3e& d\end{vmatrix}=6.$$
Finally, we multiply the first column of the matrix $\begin{pmatrix}c-2a& 3b& a\\ i-3c+6a-2g& 3h-9b& g-3a\\ f-2d& 3e& d\end{pmatrix}$ by 3 to get the matrix $\begin{pmatrix}3c-6a& 3b& a\\ 3i-9c+18a-6g& 3h-9b& g-3a\\ 3f-6d& 3e& d\end{pmatrix}$ and we have
$$\begin{vmatrix}3c-6a& 3b& a\\ 3i-9c+18a-6g& 3h-9b& g-3a\\ 3f-6d& 3e& d\end{vmatrix}=3\begin{vmatrix}c-2a& 3b& a\\ i-3c+6a-2g& 3h-9b& g-3a\\ f-2d& 3e& d\end{vmatrix}=(3)(6)=18.$$
Thank you.
 [/INST] Very well done, down to the meticulous details!  
Your solution is very easy to read and to follow; the rationale and your calculations are correct, and I couldn't find any grammatical or spelling errors.
Remark: I wish all posters were as industrious as you in taking the time to carefully format your work, not just to hand in, but also here in your post at MSE!
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