Matrix determinant effect

Yes that's correct, indeed by the properties of the determinant we have that

$$\det\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}=\det\begin{bmatrix}a&b&c\\d&e&f\\g-a&h-b&i-c\end{bmatrix}=\\=\frac13 \det\begin{bmatrix}3a&3b&3c\\d&e&f\\g-a&h-b&i-c\end{bmatrix}=-\frac13 \det\begin{bmatrix}3a&3b&3c\\-d&-e&-f\\g-a&h-b&i-c\end{bmatrix}$$

Sometimes you can't tell if matrix $B$ can be generated from matrix $B$ or not. In such a case, one may expand each determinant and compare the result of each. However, this approach requires careful attention to signs!

Let $X$ be the determinant of the first matrix and $Y$ be the determinant of the 2nd matrix, then we have:

$ X = a(ei - fh) - b(di - fg) + c(dh - eg) $

$ X = aei - afh - bdi + bfg + cdh - ceg $

$ Y = 3afh - 3eai + 3ecg + 3bdi - 3bfg - 3cdh $

$ \frac{Y}{-3} = -afh + eai -ecg -bdi + bfg + cdh $

You could re-arrange the terms of any of the equations to see that:

$X=\frac{Y}{-3}$

Since X=4,

$Y=-12$