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I have to compute the determinant of this 4x4 matrix:

\begin{bmatrix}2&1&3&0\\-1&0&1&2\\2&0&-1&-1\\-3&1&0&1\end{bmatrix}

this is what I did: I swapped first column with second column (and according to linear algebra theory, the determinant becomes negative: -det. But I swapped again in order to keep the determinant positive,therefore I swapped third column with second column. I did it, because I prefer working with "1" as the first entry of the matrix (the entry with i = j = 1).

Then, I did:

4th row = (-1) * (4th row) * (1st row) + (4th row). And I got:

\begin{bmatrix}1&3&2&0\\0&1&-1&2\\0&-1&2&-1\\0&0&3&1\end{bmatrix}

Now, I used Laplace theorem in the first column, and then Sarrus' rule. Since the first entry is always positive ( because -1 raised to an even number is always equal to 1), I used directly Sarrus' rule, as follows:

( (1 * 2 * 1) + 0 + (-6)) = -4 (this is the positive permutations)

( (0 - (-3) - 1) = 2 (this is the negative permutations)

Therefore, -4+2 = -2.

(according to Sarrus' rule, I copied the first column and the second column at the end of the minor in order to compute the determinant using its definitions (recall, it involves the positive/negative permutations) \begin{bmatrix}1&-1&2&1&-1\\-1&2&-1&-1&2\\0&3&1&0&3\end{bmatrix}

Apparently, the determinant is -2, but the exercise suggests otherwise. According the the result, the determinant of this matrix should have been equal to 15. What's wrong with this exercise?

[EDIT] - solution: I applied incorrectly row operations, because multiplying a row by a row isn't a row operation (but I can multiply a row by a multiple of the same row). So I have to subtract R1 from R4, as follows: R4 - R1.

Gabriel Burzacchini
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    Your calculation of the new 4th row is incorrect. – A.Γ. Nov 14 '21 at 09:41
  • why? I thought it was correct, could you give me a further explanation? – Gabriel Burzacchini Nov 14 '21 at 09:43
  • Ah, I think I got it, thanks. I should have written "-1 * R4 * R1 + R4", is that right? – Gabriel Burzacchini Nov 14 '21 at 09:45
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    @GabrielBurzacchini Please just check for calculation errors. The $4$th row is $(1, 0, -3, 1)$ and you are taking away the first row $(1,3,2,0)$, you should get $(0, -3, -5, 1)$ and not $(0, 0, 3, 1)$. –  Nov 14 '21 at 09:46
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    Just notice that $$ -(1\ 3\ 2\ 0)+(1\ 0\ -3\ 1)\ne(0\ 0\ 3\ 1). $$ – A.Γ. Nov 14 '21 at 09:47
  • @StinkingBishop I did -R4*R1+R4, and I got: (0, 0, 3, 1) again. I don't understand why you obtained "-3" in (0, -3, -5, 1). it should be positive, not negative, am I wrong? – Gabriel Burzacchini Nov 14 '21 at 10:09
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    To cancel $1$ in 4th row, you need to subtract first row from the fourth row, or in other words add $-1$ multiple of first row to the fourth row (as was already mentioned). So the new R4 becomes "(-1) * R1 + R4" in your notation. Why would you do "-1 * R4 * R1 + R4" instead? And what do you even mean by "R4 * R1"? – Sil Nov 14 '21 at 10:22
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    When you take away $R_1=(1,3,2,0)$ from $R_4=(1,0,-3,1)$, you get $R_4-R_1=(1-1, 0-3, -3-2, 1-0)=(0, -3, -5, 1)$. –  Nov 14 '21 at 10:27
  • @Sil I knew that multiplying a row by a scalar, and multiplying that by another row and then sum what I got by the original row, is legit (according to linear algebra theory). Is it wrong? – Gabriel Burzacchini Nov 14 '21 at 10:28
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    No, "multiplying row by another row" is not in any elementary row operation – Sil Nov 14 '21 at 10:31

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