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I have a 3x3 matrix as follows: \begin{bmatrix}4&3&1\\2&0&2\\2&1&0\end{bmatrix} I know that the correct determinant is 6, and can obtain this value using various elementary row operation routes or other methods. However, I am confused as to why the following series of operations fails to end with a determinant of 6:

Replacing row 2 with row 1 - 2*(row 2): \begin{bmatrix}4&3&1\\0&3&-3\\2&1&0\end{bmatrix}

Replacing row 3 with row 1 - 2*(row 3): \begin{bmatrix}4&3&1\\0&3&-3\\0&1&1\end{bmatrix}

Replacing row 3 with row 2 - 3*(row 3): \begin{bmatrix}4&3&1\\0&3&-3\\0&0&-6\end{bmatrix}

This gives a determinant of -72, which is way off from the correct answer of 6. As far as I can tell, I didn't use scaling at any point, nor switched any rows. I feel as though I'm missing some kind of important restriction when using elementary row operations for this purpose, or maybe just overlooking a simple detail. Any help is appreciated.

Edit: If row reduction in the manner described above changes the determinant, why does the following sequence result in 6?

Replacing row 2 with -1/2 *(row 1) + row 2

Replacing row 3 with -1/2 *(row 1) + row 3

Replacing row 3 with -1/3 *(row 2) + row 3

Would this cause the determinant to be multiplied by -1/12?

Or is it specifically related to the multiplication of the row being replaced?

Edit 2:

Thank you, I misread the answer. I'll accept it as soon as I'm allowed.

Steven Seader
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  • The determinant is not fixed by row-reduction operations. So you change the determinant at each step. Simple as that. – Mike Pierce May 15 '18 at 05:31
  • Is there a particular reason why? I was under the impression that elementary row operations changed the final determinant in the following ways: row replacement having no effect, row interchange requiring multiplying the determinant by (-1), and multiplying by r if a row is reduced by a factor of r. – Steven Seader May 15 '18 at 05:33
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    Replacing row 2 with row 1 - 2*(row 2) multiplies determinant by -2. – xyzzyz May 15 '18 at 05:34
  • You’ve in fact scaled twice. – amd May 16 '18 at 01:34

1 Answers1

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Replacing row i with row i + n*row j doesn't change the determinant. Replacing row i with row j + n*row i does (it multiplies the determinant by n).

Arthur
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  • Thank you! Can it be described (in simple terms) as having scaled row 2, then performing a row replacement? – Steven Seader May 15 '18 at 05:47
  • @StevenSeader Yes, you can decompose it into replacing row i with n*row i, which multiplies the determinant by n, and then replacing row i with row i + row j, which keeps the determinant unchanged. – Arthur May 15 '18 at 05:55
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    @StevenSeader You should look again at the definition of elementary row operations. In fact replacing row i with row j plus a multiple of row i is not an elementary row operation. – David C. Ullrich May 15 '18 at 14:37
  • @DavidC.Ullrich Yes, upon receiving an answer, I went back and looked at the definitions, which held the answer I sought. I hadn't realized that the row being replaced was supposed to remain unchanged, initially from not reading closely enough, but perpetuated by the fact that it hadn't impeded the simple row reduction I was doing up until this point. – Steven Seader May 15 '18 at 17:35
  • @StevenSeader That non-elementary row operation is in fact the composition of two elementary row operations, so there's no hard in using it in row-reducing matrices - the main place where it's dangerous, as far as I can see, is exactly here: Confusing it with the actual elementary row operation and hence assuming it does ot change the determinant. – David C. Ullrich May 15 '18 at 17:38