1
$\begingroup$

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

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.?

enter image description here

$\endgroup$
4
  • $\begingroup$ You take a matrix $A$. Apply Gaussian elimination (elementary row operations) and reduce $A$ to an upper triangular form. Then the so-called solving the determinant will be simply computing the determinant of the resulting triangular matrix. $\endgroup$
    – user2468
    Mar 4, 2012 at 22:48
  • 2
    $\begingroup$ Terminology alert: "trace" does not mean the product of the diagonal elements -- it is the sum of the diagonal elements, and is a different thing from the determinant. $\endgroup$ Mar 4, 2012 at 23:27
  • $\begingroup$ In the second step you are doing two conflicting row operations at the same time, and it seems you have gotten them mixed up. If the first operation changes the $2$ to a $0$ before subtracting row 3 from row 4, it should also change the $0$ to $\frac 83$ before subtracting that from the lower right $1$. $\endgroup$ Mar 4, 2012 at 23:30
  • $\begingroup$ @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!) $\endgroup$
    – hhh
    Mar 5, 2012 at 1:37

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ ...yes but look I can substract third row as many times as possible from the second row (i.e. I can change the diagonals so getting many different determinants which must be false), it is not unique in that way. My problem lies in this point. Could cover this point? $\endgroup$
    – hhh
    Mar 4, 2012 at 22:42
  • $\begingroup$ @hhh start with the first row and the first column. That's your pivot. Try to zero out entries in column 1 & rows 1,2. Then move to entry (2,2) that's your new pivot. Try to zero out all the entries in column 2 below (2,2). That's how Gaus.-Elim. works. $\endgroup$
    – user2468
    Mar 4, 2012 at 22:45
  • $\begingroup$ Look at the updated q, where do I do the mistake? (I left out the inverse matrix thing because I do not need it afaik) $\endgroup$
    – hhh
    Mar 4, 2012 at 22:59
  • 1
    $\begingroup$ @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. $\endgroup$
    – user2468
    Mar 4, 2012 at 23:38
0
$\begingroup$

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.
$\endgroup$
1
  • $\begingroup$ ...yes but what about with $\pm$ -additions? Look at the updated q, how do I take them into account in calculating the $det(A)$? $\endgroup$
    – hhh
    Mar 4, 2012 at 23:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .