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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) $\displaystyle l_12l_3\ \ \ and\ \ \ l_22l_3$ 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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calculate, deta, determinant, gaussian, method 
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