MHB Help with finding the determinant using gaussian elimination

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To find the determinant using Gaussian elimination for the matrix [1 2 3; 3 2 2; 0 9 8], the first step involves transforming row 2 by subtracting 3 times row 1, resulting in [1 2 3; 0 -4 -7; 0 9 8]. The next step is to modify row 3 by adding 9/4 times row 2, which leads to the matrix [1 2 3; 0 -4 -7; 0 0 -31/4]. The determinant can then be calculated as the product of the diagonal elements, yielding a final determinant value of 31. Proper execution of these row operations is crucial for accuracy in determining the determinant.
brunette15
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I attempting to find the determinant using gaussian elimination for the following matrix [1 2 3; 3 2 2; 0 9 8].
I have begun by attempting to form zeros below the diagonal. My first row operation was to make row 2 equal to 3(row 1) - row 2. This gives me [1 2 3; 0 4 7; 0 9 8] . I think i am making a small mistake whenever i try to continue from here, anyone have any suggestions as to what to do from here?
 
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brunette15 said:
I attempting to find the determinant using gaussian elimination for the following matrix [1 2 3; 3 2 2; 0 9 8].
I have begun by attempting to form zeros below the diagonal. My first row operation was to make row 2 equal to 3(row 1) - row 2. This gives me [1 2 3; 0 4 7; 0 9 8] . I think i am making a small mistake whenever i try to continue from here, anyone have any suggestions as to what to do from here?

The correct procedure starts adding to the row 2 the row 1 multiplied by -3 so that you first obtain from [1 2 3; 3 2 2; 0 9 8] the matrix [1 2 3; 0 -4 -7; 0 9 8]. Then You add to the row 3 the row 2 multiplied by 9/4 obtaining [1 2 3; 0 -4 -7; 0 0 - 31/4] , so that the determinant is 31...

Kind regards

$\chi$ $\sigma$
 
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