4.7soln - Solutions to even problems from section 4.7...

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Unformatted text preview: Solutions to even problems from section 4.7 vector(( 2. Let B = {b1 , b2 } and C = {c1 , c2 } be bases for a (hhhhh V , and suppose b1 = −c1 +4c2 (( space and b2 = 5c1 − 3c2 . a. Find the change-of-coordinates matrix from B to C . b. Find [x]C for x = 5b1 + 3b2 . Use part (a). Solution: a. Since [b1 ]C = −1 4 and [b2 ]C = 5 −3 the theorem from this section tells us −1 5 4 −3 subspace of R h ( m P = C←B b. Since [x]B = 5 3 we see that P C←B . [x]C = [x]B = −1 5 4 −3 5 3 = 10 11 m vector(( 6. Let D = {d1 , d2 , d3 } and F = {f1 , f2 , f3 } be bases for a (hhhhh (( space f1 = 2d1 − d2 + d3 , f2 = 3d2 + d3 , and f3 = −3d1 + 2d3 . a. Find the change-of-coordinates matrix from F to D . b. Find [x]D for x = f1 − 2f2 + 2f3 . Solution: 0 2 a. Since [f1 ]D = −1 , [f2 ]D = 3 , and [f3 ]D 1 1 tells us 2 P = −1 D←F 1 1 b. Since [x]F = −2 we see that 2 [x]D = P D←F subspace of R h ( V , and suppose −3 = 0 the theorem from this section 2 0 −3 3 0 . 1 2 2 0 −3 1 −4 [x]F = −1 3 0 −2 = −7 11 2 2 3 8. Let B = {b1 , b2 } and C = {c1 , c2 } be bases for R2 . Find the change-of-coordinates matrix from B to C and the change-of-coordinates matrix from C to B where b1 = −1 8 , b2 = 1 −5 , c1 = 1 4 , c2 = 1 1 . Solution: Let’s write the bi ’s as a linear combination of the ci ’s: 1 1 −1 1 41 8 −5 N R2 =− 1 R2 3 N R2 =R2 −4R1 −→ 1 1 −1 1 0 −3 12 −9 10 3 −2 0 1 −4 3 3 −4 and [b2 ]C = −2 . 3 −→ 1 1 −1 1 0 1 −4 3 N R1 =R1 −R2 −→ Thus we have b1 = 3c1 − 4c2 and b2 = −2c1 +3c2 so that [b1 ]C = Therefore P = C←B P = B←C P C←B −1 3 −2 −4 3 1 9−8 . Finally, = 32 43 = 32 43 . 10. Let B = {b1 , b2 } and C = {c1 , c2 } be bases for R2 . Find the change-of-coordinates matrix from B to C and the change-of-coordinates matrix from C to B where b1 = 7 −2 , b2 = 2 −1 , c1 = 4 1 , c2 = 5 2 . Solution: Let’s write the bi ’s as a linear combination of the ci ’s: 45 7 2 1 2 −2 −1 1 N R1 =− 3 R1 N R1 =R1 −4R2 −→ 0 −3 15 6 1 2 −2 −1 R1 ↔R2 −→ 0 1 −5 −2 1 2 −2 −1 N R2 =R2 −2R1 −→ 0 1 −5 −2 10 8 3 −→ 8 −5 10 8 3 0 1 −5 −2 and [b2 ]C = 3 . −2 Thus we have b1 = 8c1 − 5c2 and b2 = 3c1 − 2c2 so that [b1 ]C = Therefore P = C←B P = B←C P C←B −1 8 3 −5 −2 . Finally, = 1 −16 + 15 −2 −3 5 8 = 2 3 −5 −8 . ...
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This note was uploaded on 01/28/2010 for the course MATH 307 taught by Professor Axenovich during the Fall '08 term at Iowa State.

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