Is the determinant of a RREF matrix equal to the determinant of the original matrix?

Question

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.

Answer 1

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*}