Manipulating determinants multiplying two rows Is it possible to multiply two rows of a matrix to make it easier to create a determinant from? For example,
$$M=\begin{pmatrix}1 & -2 & 3\\ 2 & 3 & -4 \\ -3 & 1 & 4\end{pmatrix}$$
$$\underline{R_3}+R_2\to \begin{pmatrix}1 & -2 & 3\\ 2 & 3 & -4 \\ -1 & 4 & 0\end{pmatrix}$$
$$\underline{R_1}\times R_3\to \begin{pmatrix}-1 & -8 & 0\\ 2 & 3 & -4 \\ -1 & 4 & 0\end{pmatrix}$$
I don't get the correct value for the determinant (which is $-41$) but get $48$ instead using the $R_1\times R_3$ matrix, yet get the correct answer on the $R_3+R_2$ matrix, am I doing something mathematically wrong multiplying two rows? 
 A: Yes, you are doing something wrong when multiplying two rows together. Consider a simpler example : you know that
$$\det\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}=1,$$
but if you multiply the first row by the second, you will get
$$\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix},$$which has determinant $0$.
The only operation you can do without changing the determinant is adding a multiple of a row to another row. Some operations change the determinant in a predictable way: if you multiply a row by some scalar $\alpha$, then the determinant gets multiplied by $\alpha$ as well, and if you switch two rows the determinant changes sign. But multiplying two rows together doesn't obey any rule (as far as I know).
A: Yes, it's mathematically wrong. There's no reason to expect that it would work as you wish. The reason is that the determinant is a linear function of each of its rows. When you replace one row by its sum with another row, the linearity tells you the determinant is the sum of the determinant of the original matrix plus the determinant of a matrix with two identical rows. That second determinant is $0$.
(That argument should appear somewhere in your linear algebra lectures or text.)
There's nothing about determinants that lets you make this argument or any other for replacing a row by its product (element by element) with another row.
