# week03 - 4 23 r 3 → r 3-2 r 4 →-1 1 2 3 1 1-21 = U r 4...

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Week 3: 1. 1.5 Determinants 2. 1.6 Properties of Dets 3. 2.1 Vector Spaces ——– We compute det 2 1 5 4 2 3 9 5 1 using the (ﬁrst) row expansion (by minors): det ( A ) = 2 2 3 5 1 - 1 4 3 9 1 + 5 4 2 9 5 = 2(2 - 15) - (4 - 27) + 5(20 - 18) = 2( - 13) - ( - 23) + 5(2) = - 26 + 33 = 7 . ———– Problem Reduce A = 2 1 3 5 3 0 1 2 4 1 4 3 5 2 5 3 to an upper triagular matrix and use the reduction to ﬁnd det ( A ) . Solution: A - 1 1 2 3 3 0 1 2 4 1 4 3 5 2 5 3 ( r 1 r 1 - r 2 ) - 1 1 2 3 0 3 7 11 0 5 12 15 0 7 15 18 ( r 2 r 2 - 3 r 1 , r 3 r 3 - 4 r 1 , r 4 r 4 - 5 r 1 )

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2 - 1 1 2 3 0 3 7 11 0 5 12 15 0 1 1 - 4 ( r 4 r 4 - 2 r 2 ) ——– - 1 1 2 3 0 1 1 - 4 0 5 12 15 0 3 7 11 ( r 2 r 4 ) - 1 1 2 3 0 1 1 - 4 0 0 7 35 0 0 4 23 ( r 3 r 3 - 5 r 2 , r 4 r 4 - 3 r 2) - 1 1 2 3 0 1 1
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Unformatted text preview: -4-1-11 4 23 ( r 3 → r 3-2 r 4 ) → -1 1 2 3 1 1-4-1-11-21 = U ( r 4 → r 4 + 4 r 3 ). Now we have det U = (-1)(1)(-1)(-21) =-21 , and det A = (-1) det U = 21 , since all of the 3rd elementary operations change the determinant by a factor of (1) – so no change - there are no 2nd EROs and exactly one 1st ERO, with each 1st ERO changing the determinant by a factor of (-1) ....
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week03 - 4 23 r 3 → r 3-2 r 4 →-1 1 2 3 1 1-21 = U r 4...

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