Linear Algebra 1 - 1 Math 2602 M1/M2 Spring 2007 Elementary...

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Math 2602 M1/M2 Spring 2007 Elementary Row Operations 1 April 3, 2007 Problem # 1 . Let B = 0 0 - 2 0 - 2 0 - 2 0 0 . Find B - 4 . Solution. Note that B - 4 = ( B - 1 ) 4 = ( B 4 ) - 1 There are two alternate ways of finding B - 4 . 1 st way. Firstly compute B - 1 by multiplying row 1, row 2, and row 3 by - 1 / 2 in [ B | I 3 ] and then reverse the order of rows so as to get I 3 on the left. This way the desired B - 1 is obtained on the right as B - 1 = 0 0 - 1 / 2 0 - 1 / 2 0 - 1 / 2 0 0 . and so B - 2 = 1 / 4 0 0 0 1 / 4 0 0 0 1 / 4 A B - 4 = 1 / 16 0 0 0 1 / 16 0 0 0 1 / 16 . 2 nd way. We will firstly compute B 4 and then ( B 4 ) - 1 . Observe that B 2 = 0 0 - 2 0 - 2 0 - 2 0 0 0 0 - 2 0 - 2 0 - 2 0 0 = 4 0 0 0 4 0 0 0 4 = 4 I 3 Therefore, B 4 = ( B 2 )( B 2 ) = (4 I 3 )(4 I 3 ) = 16 I 2 3 = 16 I 3 = 16 0 0 0 16 0 0 0 16 Remember that for any a 6 = 0 , b 6 = 0 , and c 6 = 0 a 0 0 0 b 0 0 0 c - 1 = 1 /a 0 0 0 1 /b 0 0 0 1 /c . As a result,
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This note was uploaded on 05/12/2010 for the course MATH 2602 taught by Professor Costello during the Spring '09 term at Georgia Institute of Technology.

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Linear Algebra 1 - 1 Math 2602 M1/M2 Spring 2007 Elementary...

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