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$A=\begin{pmatrix} 1 & -1 & 0 & 2 \\ 2 & 1 & 0 & 0 \\ 1 & 1 & 2 & 2 \\ 0 & 0 & 1 & 1 \\ \end{pmatrix}$

With the lower determinant method, I got $det(A)=-2$ but my task is to use Gauss method to find out determinant. I know that for a triangular matrix $B$, $det(B)=\prod b_{ii}$ i.e. the trace (product of diagonal things). Now I can make this into a triangular matrix by Gauss Jordan but I cannot understand yet what does it mean that solve the determinant with Gauss method or Gauss rule whatever you call it? I am on page 741 XI.5:4, here (not English), it should be trivial problem but stuck to this.

ERR: what is the problem with this, trying to use the G.E.?

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    @HenningMakholm: thanks, clearly it is "true" now that it is hard to do two things at the same time. Thanks for the notice. Irritating mistake. (the latter thing about truth was meant to be self-irony --- how easy problem with enough eye-balls!)2012-03-05

2 Answers 2

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Performing Gaussian Elimination on $ A=\begin{pmatrix} 1 & -1 & 0 & 2 \\ 2 & 1 & 0 & 0 \\ 1 & 1 & 2 & 2 \\ 0 & 0 & 1 & 1 \\ \end{pmatrix}$ we get $ \begin{pmatrix} 1 & -1 & 0 & 2 \\ 0 & ? & ? & ? \\ 0 & 0 & ? & ? \\ 0 & 0 & 0 & ? \\ \end{pmatrix}$ which has the determinant (see property 9 here): $ 1 \times ?\times ? \times ? = -2 $ Since this is a (homework) question, I will let you fill in all the blanks.

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    @hhh First matrix is okay. Second matrix is okay. Transition from 2nd matrix to 3rd matrix is wrong. Do only one row operation, namely to zero out (3,2). Your 3rd matrix should have the 4th row: $[0, 0, 1, 1]$. Then to get the 4th matrix, add $(-1/2)*[0,0,2,8/3]$ to $[0,0,1,1]$. That should get you exactly what you want. **Also note that as Henning pointed out, trace is the sum of the diagonal entries. The determinant as in my solution above is the product of the diagonal entries.**2012-03-04
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I think this might help: Det via Gaussian Elimination. If you have computed the triangular matrix using Gauss Jordan, all you need is to be careful with the following 2 rules while applying the elimination:

  1. Interchanging two rows changes the sign of the determinant.
  2. Multiplying a row by a scalar multiplies the determinant by that scalar.
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    ...yes but what about with $\pm$ -additions? Look at the updated q, how do I take them into account in calculating the $det(A)$?2012-03-04