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Unformatted text preview: Example 5. Use cofactor expansion along the 3 rd row to calculate det A where A = 12 1 2 1 211 1 2 1 1 Since a 31 = a 32 = a 34 = 0 there is only one nonzero term det A = 0 + 0 + (1) · A 33 + 0 where A 33 = (+) · det 12 2 11 1 2 1 1 Since it’s a 3 × 3 determinant we can compute this fairly easily to be 1 + 2 + 0(2)(4) = 9, so the ﬁnal answer is det A = a 33 A 33 = (1) · 9 =9 MATLAB agrees: >>> det([1,2,1,0;2,1,2,1;0,0,1,0;1,2,1,1]) ans = 9 2...
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 Spring '08
 Bens
 Linear Algebra, Algebra, Determinant, Howard Staunton, Mij, anc Anc

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