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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.

Rodrigo de Azevedo
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Cristiano Ronaldo
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  • Show that the $\det$ of the EROs all have determinant 1. – copper.hat Mar 24 '15 at 05:04
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    Either I'm missing something or this is not true. If $A$ is an invertible square matrix, its r.r.e.f. $R$ is $I$, which has determinant $1$, but such $A$ need not have determinant $1$. On the other hand, each of the row reduction operations modifies the determinant of a matrix in a simple way, so one can easily compute the determinant by tracing these modifications through. – Travis Willse Mar 24 '15 at 05:06
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    @copper, isn't multiplying a row by 17 an ERO? – Gerry Myerson Mar 24 '15 at 05:57
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    @GerryMyerson: Well, $17 \approx 1$... – copper.hat Mar 24 '15 at 06:10
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    EROs (hmm?) come in three varieties: (i) Row switching ($\det E= -1$), (ii) Row multiplication ($\det E = \alpha$, where $\alpha$ is the row multiplier). (iii) Row addition ($\det E = 1$). – copper.hat Mar 24 '15 at 06:16
  • It is straightforward to see that either $\det R = 0$ or $\det R = 1$. Hence it is straightforward to show that, in general, $\det R \neq \det A$. Take the $1 \times 1$ matrix $A=2$, for example. The reduced row echelon form of $A$ is $1$. – copper.hat Mar 24 '15 at 06:18

2 Answers2

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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.

Rodrigo de Azevedo
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zthomas.nc
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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*}

itp_tk
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