My problem:
Suppose $$ \begin{vmatrix}a & b& c \\ d & e & f \\ g& h & i\end{vmatrix} = 8. $$ Then find the value of $$ \begin{vmatrix} (g+2a) & (h+2b)& (i+2c) \\ 3a & 3b & 3c \\ 2d& 2e & 2f\end{vmatrix}. $$
I believe the final answer is $-48$ but I'm not sure.
I hate having to open a link when it is not that difficult to type the problem here! The problem is "Given that $\left|\begin{array}{ccc}a & b & c \\ d & e & f \\ g & h & i\end{array}\right|= 8$
Find the value of $\left|\begin{array}{ccc}(g+ 2a) & (h+ 2b) & (i+ 2c) \\ 3a & 3b & 3c \\ 2d & 2c & 2f\end{array}\right|$."
There are three kinds of "row operations": 1) Multiply a row by a number. This multiplies the determinant by the number. 2) Swap two rows. This multiplies the determinant by -1. 3) Add a multiple of one row to another. This does not change the determinant.
Starting from $\left|\begin{array}{ccc}a & b & c \\ d & e & f \\ g & h & i\end{array}\right|$, multiply the first row by 2 to get $\left|\begin{array}{ccc}2a & 2b & 2c \\ d & e & f \\ g & h & i\end{array}\right|= 2(8)= 16$.
Add the third row to the first row of that to get $\left|\begin{array}{ccc}g+ 2a & h+ 2b & i+ 2c \\ d & e & f \\ g & h & i\end{array}\right|= 16$.
Multiply the second row of that by 3 to get $\left|\begin{array}{ccc}g+ 2a & h+ 2b & i+ 2c \\ 3d & 3e & 3f \\ g & h & i\end{array}\right|= 3(16)= 48$.
Finally, multiply the third row of that by 2 to get $\left|\begin{array}{ccc}g+ 2a & h+ 2b & i+ 2c \\ 3d & 3e & 3f \\ 2g & 2h & 2i\end{array}\right|= 2(48)= 96$.