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Linear algebra determinant 
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#1
Oct2410, 03:58 PM

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1. The problem statement, all variables and given/known data
I am finding the determinant by first reducing to row echelon form. Please help I have been agonizing over this lol 2. Relevant equations I have got to this stage and am now going to do 3rd row  4th row and replace the 4th row as you do to get to row echelon. At this stage the determinant is 0.8 1 3 2 1 0 1 2 1 0 0 1 0.8 0 0 1 0 Now I obtain this matrix and since it is now a triangular matrix, the determinant is now 0.8. Yet I have simply added a multiple of one row to another which is meant to leave the determinant unchanged. So what on earth have I done wrong? Thanks for answers. 1  3  2  1 0  1  2  1 0  0  1  0.8 0  0  0  0.8 3. The attempt at a solution 


#2
Oct2410, 04:38 PM

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from your other thread on the same problem …



#3
Oct2410, 04:44 PM

P: 14

Look I'm sorry I don't know what he means. Look at the second matrix, the determinant must be 0.8. I think he means replacing the 3rd row but with row echelon form I have to replace the 4th row



#4
Oct2410, 04:52 PM

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Linear algebra determinant



#5
Oct2410, 04:54 PM

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#6
Oct2410, 05:10 PM

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Your bottom righthand corner (depending on which row you subtract from the other) will be either 1 0.8 0 0.8 or 0 0.8 1 0 both of which have determinant 0.8 


#7
Oct2410, 05:10 PM

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#8
Oct2410, 05:17 PM

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1 0.8 0 0.8 


#9
Oct2410, 05:30 PM

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You can replace the 4th row by adding the 3rd row to the 4th row. You can't muck about with the row you're replacing (you've multiplied it by 1 ) before you start adding to or subtracting other rows from it. 


#10
Oct2410, 05:39 PM

P: 14

Is this correct? Thanks for the help 


#11
Oct2410, 05:47 PM

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But you're adding (or subtracting) the wrong row. Look at your book again. 


#12
Oct2410, 05:57 PM

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#13
Oct2510, 02:53 AM

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(just got up …)
some techniques work and some don't … there are theorems in books which tell you which techniques work and why you really do need to go back to your book and study this again, from the beginning … 


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