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Suppose you are trying to find the determinant of the following matrix using the "upper triangulation" method:

$\begin{matrix} 1&0&0\\ 0&1&0\\ 1&1&1 \end{matrix}$

If I take R1 - R3 -> R3 (row 1 minus row 3 and put the result in row 3):

$\begin{matrix} 1&0&0\\ 0&1&0\\ 0&-1&-1 \end{matrix}$

Then I do R2 + R3 ->R3:

$\begin{matrix} 1&0&0\\ 0&1&0\\ 0&0&-1 \end{matrix}$

And the determinant is then = -1 (because I multiply the elements in the diagonal)

However if the first step is instead:

R3 - R1 ->R3:

$\begin{matrix} 1&0&0\\ 0&1&0\\ 0&1&1 \end{matrix}$

And then R3 - R2 -> R3:

$\begin{matrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{matrix}$

The determinant is 1 !!!

How do you know which row is subtracted from what?

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  • $\begingroup$ Somewhere you interchanged two rows, which means you change the orientation so then the determinant changed sign. $\endgroup$ – Kaladin Mar 8 '14 at 21:19
  • $\begingroup$ your mistake is (if i understand you correctly): if $range\left(A\right)$ = $range\left(B\right)$ that doesn't mean that $det\left(A\right)=det\left(B\right)$. (the line operations you apply are keeping the range constant) $\endgroup$ – Max Mar 8 '14 at 21:20
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You can add or remove any row to any other (not the same) row, but in the first step you replace R3 by -R3 (+R1 but you can add). So by negating a row, you negate the determinant.

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  • $\begingroup$ Ah, so whenever you are subtracting a row from another row, you essentially are multiplying the determinant by -1? $\endgroup$ – W. Heisenberg Mar 8 '14 at 21:34
  • $\begingroup$ No, adding a scalar multiple of one row to another does not affect the determinant. Interchanging rows however, has the same affect as multiplying det by -1. And multiplying a row by a scalar a is the same as multiplying the determinant by a. $\endgroup$ – Zachary Mar 8 '14 at 21:44

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