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Unformatted text preview: i24 i ± ± ± ± ± ± R 13 * R 2 = (1i ) ± ± ± ± ± ± 4 + 3 i 22 i 1i i 1 1i24 i ± ± ± ± ± ± = (1i )(1) ± ± ± ± ²4 + 3 i 22 i 1i24 i ³± ± ± ± = (1i )(1)[(4 + 3 i )(24 i )(1i )(22 i )] = (1 + i )[(8 + 12 + 16 i6 i )(222 i2 i )] = (1 + i )(20 + 14 i ) = 20 + 14 i + 20 i14 = 6 + 34 i ± Note: I hope that the calculations are correct. If you ﬁnd any mistakes, please, let me know. Thanks. 2...
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This note was uploaded on 11/30/2009 for the course MATH 115A taught by Professor Liu during the Winter '07 term at UCLA.
 Winter '07
 Liu
 Linear Algebra, Algebra

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