Is the determinant of a RREF matrix equal to the determinant of the original matrix? Prove or disprove: If $R$ is the reduced row echelon form (RREF) of $A$, then $\det A = \det R$, where $A$ is an $n \times n$ matrix.
 A: Well, I'm not sure about "proving" this, but a contradiction to your statement:

*

*If $A = \begin{bmatrix}
17 & 0 \\
0 & 17 
\end{bmatrix}$, then $\det(A)$ = $17^2$


*If $B = \begin{bmatrix}
1 & 0 \\
0 & 1 
\end{bmatrix}$, then $\det(A)$ = $1^2$  = 1
Also, $B$ is the reduced echelon form of $A$. Therefore, the determinant of a matrix and its reduced echelon form is not necessarily the same.
A: No you can't simply apply directly the properties that you use with matrices, as @zthomas.nc proved it wrong:

If A = $\begin{bmatrix}
17 & 0 \\
0 & 17 
\end{bmatrix}$, then $det(A)$ = $17^2$


If B = $\begin{bmatrix}
1 & 0 \\
0 & 1 
\end{bmatrix}$, then $det(A)$ = $1^2$  = 1.

You can use determinants properties to get sort of "echelon-determinant".
1. Linearity of determinant
Given a $n \times n$ matrix $A$ we indicate its columns as $C_1, C_2, C_3, \dots, C_n$ or its rows as $R_1, R_2, R_3, \dots, R_n$.
Linearity says, for both columns and rows:

*

*$\textrm{det}(C_1, C_2, \dots, \lambda C_i, \dots, C_n) = \lambda\textrm{det}(C_1, C_2, \dots, C_i, \dots, C_n)$ (homogeneity);

*$\textrm{det}(C_1, C_2, \dots, C_i + \tilde C_i, \dots, C_n) = \textrm{det}(C_1, C_2, \dots, C_i, \dots, C_n) + \textrm{det}(C_1, C_2, \dots, \tilde C_i, \dots, C_n)$ (additivity).

Note: A consequence of the homogeneity property is
\begin{equation*}
\textrm{det}(\lambda A) = \lambda^n\textrm{det}(A).
\end{equation*}
Note: If two columns or rows are proportional the determinant is zero
\begin{equation*}
\textrm{det}(C_1, C_2, \dots, C_i, \dots, \lambda C_i, \dots, C_n) = \lambda\textrm{det}(C_1, C_2, \dots, C_i, \dots, C_i, \dots, C_n) = 0
\end{equation*}
2. Linear combination of rows or columns
The determinant of $A$ doesn't change if we sum to a column or row a linear combination of other columns or rows
\begin{equation*}
\textrm{det}(C_1, C_2, \dots, C_i, \dots, C_j \dots, C_n) = \textrm{det}(C_1, C_2, \dots, C_i + \lambda C_j, \dots, C_j, \dots, C_n) = \textrm{det}(C_1, C_2, \dots, C_i, \dots, C_j, \dots, C_n) + \textrm{det}(C_1, C_2, \dots, \lambda C_j, \dots, C_j, \dots, C_n) = \textrm{det}(C_1, C_2, \dots, C_i, \dots, C_j, \dots, C_n) + 0.
\end{equation*}
3. Echelon form
Given this two properties we can now find a sort of echelon form, in the sense that we can sum linear combination of rows and columns and use the linearity to have a simplified form of the determinant.
So the previous example A = $\begin{bmatrix}
17 & 0 \\
0 & 17 
\end{bmatrix}$ would be
\begin{equation*}
\left\lvert \begin{matrix}
17 & 0 \\
0 & 17 \\
\end{matrix} \right\rvert = 17^2 \left\lvert \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right\rvert = 17^2 = 289.
\end{equation*}
