Unformatted text preview: ) 1 a a 2 1 b b 2 1 c c 2 → 1 a a 2 ba b 2a 2 ca c 2a 2 → 1 a a 2 ba b 2a 2 c 2a 2ca ba ( b 2a 2 ) . The last entry becomes c 2a 2( ca )( b + a ) = ( ca )(( c + a )( b + a )) = ( ca )( cb ) so det ( A ) = ( ca )( cb )( ba ). 24 ) det ( L ) = 1 3 = 1, det ( U ) = (3)(2)(1) =6, det ( A ) = det ( L ) det ( U ) =6. As U1 L1 = A1 , det ( U1 L1 ) = 1 /det ( A ) =1 / 6, and U1 L1 A = I so it has determinant 1....
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 '05
 HUI
 Math, Linear Algebra, Algebra, Prime number, 1 L, Det

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