My Math Forum Calculate det(a) using Gaussian Method

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 December 29th, 2014, 04:54 AM #1 Newbie     Joined: Dec 2014 From: Poland Posts: 1 Thanks: 0 Calculate det(a) using Gaussian Method I don't understand how to get 33 as det of A |2 3 -2 4| |2 4 -3 2| |1 3 2 1| |0 1 2 3| I tried to swap row 3&4 to 1&2 and get my echelon lower bound triangles for I get det(A)= 1 I tried without swaping, step by step zero each rowxcolumn but I end up with det(a)= 22 Could somebody help me out, how to tackle this? thx in advance!
 December 29th, 2014, 11:20 AM #2 Senior Member     Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177 $\displaystyle l$ = line (row). I) $\displaystyle l_1-2l_3\ \ \ and\ \ \ l_2-2l_3$ From here, can get easy $\displaystyle det(A) = \begin{vmatrix}-3\ \ -6\ \ \ 2 \\\;\\ -2\ \ -7\ \ \ 0 \\\;\\ \ \ 1 \qquad 2 \ \ \ \ 3 \end{vmatrix}$ II) $\displaystyle l_1+3l_3\ \ \ and\ \ \ l_2+2l_3$ Then, we get det(A) =$\displaystyle \begin{vmatrix}0 \ \ \ 0 \ \ \ 11 \\\;\\ \ 0 \ -3\ \ \ 6 \\\;\\ 1 \quad 2 \ \ \ \ 3 \end{vmatrix}\ =\ 33.$
 January 19th, 2015, 11:12 PM #3 Newbie   Joined: Jan 2015 From: pakistan Posts: 1 Thanks: 0 I tried to swap row 3&4 to 1&2 and get my echelon lower bound triangles for I get det(A)= 1 I tried without swaping, step by step zero each rowxcolumn but I end up with det(a)= 22
 January 20th, 2015, 07:03 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 You have $\displaystyle \left|\begin{array}{cccc} 2 & 3 & -2 & 4 \\ 2 & 4 & -3 & 2 \\ 1 & 3 & 2 & 1 \\ 0 & 1 & 2 & 3\end{array}\right|$ Yes, one thing you could do is swap rows 1 and 3 and rows 2 and 4. That would give $\displaystyle \left|\begin{array}{cccc} 1 & 3 & 2 & 1 \\ 0 & 1 & 2 & 3\\ 2 & 3 & -2 & 4 \\ 2 & 4 & -3 & 2 \end{array}\right|$ which has "1" and "0" for the first two numbers in the first column. Further, since swapping two rows multiplies the determinant by -1 and you have swapped twice, you haven't changed the determinant. Now, subtract twice the first row from both third and fourth rows: $\displaystyle \left|\begin{array}{cccc} 1 & 3 & 2 & 1 \\ 0 & 1 & 2 & 3\\ 0 & -3 & -6 & 2 \\ 0 & -2 & -7 & 0 \end{array}\right|$ so the first column is finished. And adding a multiple of one row to another does not change the determinant. Now, add 3 times the second row to the third and add 2 times the second row to the fourth: $\displaystyle \left|\begin{array}{cccc} 1 & 3 & 2 & 1 \\ 0 & 1 & 2 & 3\\ 0 & 0 & 0 & 11 \\ 0 & 0 & -3 & 6\end{array}\right|$ Since we have a "0" in the third column of the third row, we will need two swap the third and fourth rows: $\displaystyle \left|\begin{array}{cccc} 1 & 3 & 2 & 1 \\ 0 & 1 & 2 & 3\\ 0 & 0 & -3 & 2\\ 0 & 0 & 0 & 11 \end{array}\right|$ Swapping two rows multiplies the determinant by -1. Since we now have an "upper triangular matrix" its determinant is the product of the numbers on the main diagonal: 1(1)(-3)(11)= -33. Since we have multiplied by -1, the determinant of the original matrix is 33.

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