101st_tut4sol

101st_tut4sol - MATH1111/2010-11/Tutorial IV Solution 1...

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MATH1111/2010-11/Tutorial IV Solution 1 Tutorial IV Suggested Solution 1. Given three ordered bases for R 3 : E = [ e 1 ; e 2 ; e 3 ], F = [ f 1 ; f 2 ; f 3 ] and H = [ h 1 ; h 2 ; h 3 ], where e 1 ; e 2 ; e 3 form the standard basis, and f 1 = (1 1 1) T ; f 2 = (1 2 2) T ; f 3 = (2 3 4) T ; h 1 = (3 2 5) T ; h 2 = (1 1 2) T ; h 3 = (2 3 2) T : (a) If [ x ] F = (1 ± 1 2) T , ±nd [ x ] E and [ x ] H . (b) Find, by evaluating [ f 1 ] H , [ f 2 ] H , [ f 3 ] H , the transition matrix from F to H . (c) Find the transition matrix from F to H by ±rst evaluating the transition matrices from F to E and from H to E . Ans . (a) From [ x ] F = (1 ± 1 2) T , we get x = f 1 ± f 2 + 2 f 3 = (1 1 1) T ± (1 2 2) T + 2(2 3 4) T = (4 5 7) T : ) x = 4 e 1 + 5 e 2 + 7 e 3 , and [ x ] E = (4 5 7) T . To ±nd [ x ] H , we need to evaluate ±;²;³ 2 R so that ± h 1 + ² h 2 + ³ h 3 = x , i.e. ± (3 2 5) T + ² (1 1 2) T + ³ (2 3 2) T = (2 5 7) T 0 @ 3 1 2 2 1 3 5 2 2 1 A 0 @ ± ² ³ 1 A = 0 @ 4 5 7 1 A : The inverse of 0 @ 3 1 2 2 1 3 5 2 2 1 A is 1 3 0 @ 4 ± 2 ± 1 ± 11 4 5 1 1 ± 1 1 A . Thus [ x ] H = ( ± 1 3 11 3 2 3 ) T . (b) By the work in Part (a), we see that [ f 1 ] H = 1 3 0 @ 4 ± 2 ± 1 ± 11 4 5 1 1 ± 1 1 A 0 @ 1 1 1 1 A = 0 B @ 1 3 ± 2 3 1 3 1 C A ; [ f 2 ] H = 1 3 0 @ 4 ± 2 ± 1 ± 11 4 5 1 1 ± 1 1 A 0 @ 1 2 2 1 A = 0 B @ ± 2 3 7 3 1 3 1 C A ; [ f 3 ] H = 1 3 0 @ 4 ± 2 ± 1 ± 11 4 5 1 1 ± 1 1 A 0 @ 2 3 4 1 A = 0 B @ ± 2 3 10 3 1 3 1 C A :
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MATH1111/2010-11/Tutorial IV Solution 2 The transition matrix from F to H is 1 3 0 @ 1 ± 2 ± 2 ± 2 7 10 1 1 1 1 A .
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This note was uploaded on 05/04/2011 for the course MATH 1111 taught by Professor Dr,li during the Spring '10 term at HKU.

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101st_tut4sol - MATH1111/2010-11/Tutorial IV Solution 1...

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