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Unformatted text preview: det = (1(2))(2(2))(3(2))(2(1))(3(1))(32) = 240 . (b) det = det ± 1 2 5 6 ² · det ± 9 10 11 12 ² = (4)(2) = 8 . (c) The second raw is 0. det = 0. (d) The determinant of an upper triangular matrix is the product of its diagonal entries. det = 12. (e) The fourth column is a multiple of the ﬁrst column. det = 0. 1 3. Solve the given system of linear equation using Cramer’s rule. x 1x 2 + 4 x 3 =28 x 1 + 3 x 2 + x 3 = 0 2 x 1x 2 + x 3 = 6 (Solution) Let A = 11 48 3 1 21 1 . Then x 1 = 1 det( A ) det 21 4 3 1 61 1 =43 , x 2 = 1 det( A ) det 12 48 1 2 6 1 =109 , x 3 = 1 det( A ) det 1128 3 21 6 =17 . Therefore ( x 1 ,x 2 ,x 3 ) = (43 ,109 ,17). 2...
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This note was uploaded on 10/02/2009 for the course MATH 54554 taught by Professor Holtz during the Spring '06 term at Berkeley.
 Spring '06
 Holtz
 Linear Algebra, Algebra

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