Determine the value of a second determinant based on the first I know the theory of determinants, but I have no idea how to apply it to this problem.
Suppose $$\det\begin{bmatrix}a&b&c\\ d&e&f\\ g&h&i \end{bmatrix} = 6$$
What is the value of
$$\det\begin{bmatrix}g - 2a&h - 2b&i - 2c\\ 3a&3b&3c\\2d&2e&2f \end{bmatrix}$$
The options for the answers are:


*

*4

*36

*24

*-24

 A: $$\det\begin{bmatrix}g - 2a&h - 2b&i - 2c\\ 3a&3b&3c\\2d&2e&2f \end{bmatrix}$$
$$=\det\begin{bmatrix}g&h&i\\ 3a&3b&3c\\2d&2e&2f \end{bmatrix}$$ (Applying $R_1'=R_1+\frac23 R_2$)
$$=3\cdot2\cdot \det\begin{bmatrix}g&h&i\\ a&b&c\\d&e&f \end{bmatrix}$$ (Taking out $3,2$ as common factors from the $R_2,R_3$ respectively)
$$=(-1)3\cdot2\cdot \det\begin{bmatrix}a&b&c\\ g&h&i\\d&e&f \end{bmatrix}$$ (Exchanging $R_1,R_2$ resulting '-' sign )
$$=(-1)(-1)3\cdot2\cdot \det\begin{bmatrix}a&b&c\\d&e&f\\ g&h&i \end{bmatrix}$$
(Exchanging $R_2,R_3$ resulting '-' sign  again)
A: We switch the first and the second row and the second and third row so
$$\det\begin{bmatrix}g - 2a&h - 2b&i - 2c\\ 3a&3b&3c\\2d&2e&2f \end{bmatrix}=3\times 2\times \det\begin{bmatrix}a&b&c\\ d&e&f\\ g - 2a&h - 2b&i - 2c\end{bmatrix}$$
and we add $2\times$ the first row to the third row we find
$$\det\begin{bmatrix}g - 2a&h - 2b&i - 2c\\ 3a&3b&3c\\2d&2e&2f \end{bmatrix}=3\times 2\times \det\begin{bmatrix}a&b&c\\ d&e&f\\ g &h &i \end{bmatrix}=6\times 6=36$$
A: det$\begin{bmatrix}g - 2a&h - 2b&i - 2c\\ 3a&3b&3c\\2d&2e&2f \end{bmatrix}$ =3$\times 2\times \det\begin{bmatrix}g - 2a&h - 2b&i - 2c\\ a&b&c\\d&e&f \end{bmatrix}$  =6$\times -1\times \det\begin{bmatrix}a&b&c\\g - 2a&h - 2b&i - 2c \\d&e&f \end{bmatrix}$R1 ->R2, R2 ->R1 = 6$\times -1\times -1\times\det\begin{bmatrix}a&b&c\\d&e&f \\g - 2a&h - 2b&i - 2c \end{bmatrix}$R2 ->R3, R3 ->R2 = 6$\times\det\begin{bmatrix}a&b&c\\d&e&f \\g - 2a&h - 2b&i - 2c \end{bmatrix}$ = 6$\times\det\begin{bmatrix}a&b&c\\d&e&f \\g&h&i\end{bmatrix}$ R3 -> 2R1+R3 As, $\det\begin{bmatrix}a&b&c\\d&e&f \\g&h&i\end{bmatrix}$=6
 = 6$\times6=36$ 
A: Also, for the fast way not the theoretical approach, you can set $a=2,e=3,i=1$ and other arrays to be zero and just examine the value of determinant you're asked. So, it gives you $36$.
A: determined of square matrix is the multiplication of all term present on diagonal. so you can set $a,e,i$ such that is gives $6$ and all other terms $0$. Now put all values in second matrix and evaluate the result it is $36$.
let $a=1, e=1,i=6$ 
$$\det\begin{bmatrix} -2&0&6\\ 3&0&0\\0&2&0\end{bmatrix}$$
it gives $36$
