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Unformatted text preview: MA 35100 HOMEWORK ASSIGNMENT #7 SOLUTIONS Problem 1. pg. 147; prob. 17 In Exercises 1 through 18, determine whether the vector ~x is in the span V of the vectors ~v 1 ,...,~v m (proceed “by inspection” if possible, and use the reduced rowechelon form is necessary). If ~x is in V , find the coordinates of ~x with respect to the basis B = ( ~v 1 ,...,~v m ) of V , and write the coordinate vector ~x B . ~x = 1 1 1 1 ; ~v 1 = 1 2 , ~v 2 = 1 3 , ~v 3 = 4 1 . Solution: First observe the identity c 1 ~v 1 + c 2 ~v 2 + c 3 ~v 3 = c 1 c 2 2 c 1 + 3 c 2 + 4 c 3 c 3 . Consider c 1 = 1, c 2 = 1, and c 3 = 1. We see that 2 c 1 + 3 c 2 + 4 c 3 = 1, so that ~x = ~v 1 + ~v 2 ~v 3 . Hence yes, ~x ∈ V . Moreover, ~x = 1 1 1 . Problem 2. pg. 147; prob. 19 In Exercises 19 through 24, find the matrix B of the linear transformation T ( ~x ) = A~x with respect to the basis B = ( ~v 1 ,~v 2 ). For practice, solve each problem in three ways: (a) Use the formula B = S 1 AS (b) use a commutative diagram (as in Examples 3 and 4), and (c) construct B “column by column.” A = 0 1 1 0 ; ~v 1 = 1 1 , ~v 2 = 1 1 . Solution: (a) The matrix S and its inverse are S = ~v 1 ~v 2 = 1 1 1 1 , S 1 = 1 2 1 1 1 1 = 1 2 1 1 1 1 . We then have the product S 1 AS = 1 2 1 1 1 1 0 1 1 0 1 1 1 1 = 1 2 1 1 1 1 1 1 1 1 = 1 2 2 2 . 1 We conclude that B = S 1 AS = 1 1 . (b) We consider a diagram in the form R 2 A→ R 2 S x y S 1 R 2 B→ R 2 where we explicitly map ~x = c 1 ~v 1 + c 2 ~v 2 = c 1 + c 2 c 1 c 2 A = 0 1 1 0 → T ( ~x ) = 0 1 1 0 c 1 + c 2 c 1 c 2 = c 1 c 2 c 1 + c 2 = c 1 ~v 1 c 2 ~v 2 x y ~x B = c 1 c 2 B = 1 1 → T ( ~x ) B = c 1 c 2 (c) To construct B “column by column” we use Definition 3.4.3 on page 143: T ( ~v 1 ) = 0 1 1 0 1 1 = 1 1 = ~v 1 , T ( ~v 2 ) = 0 1 1 0 1 1 = 1 1 = ~v 2 ; which gives T ( ~v 1 ) B = 1 and T ( ~v 2 ) B = 1 . We then find B = [ T ( ~v 1 )] B [ T ( ~v 2 )] B = 1 1 . Problem 3. pg. 147; prob. 20 In Exercises 19 through 24, find the matrix B of the linear transformation T ( ~x ) = A~x with respect to the basis B = ( ~v 1 ,~v 2 ). For practice, solve each problem in three ways: (a) Use the formula B = S 1 AS (b) use a commutative diagram (as in Examples 3 and 4), and (c) construct B “column by column.”“column by column....
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This note was uploaded on 10/18/2010 for the course MATH 351 taught by Professor Egoins during the Spring '10 term at Purdue.
 Spring '10
 EGoins
 Linear Algebra, Algebra, Vectors

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