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Unformatted text preview: Chapter 2 SSM: Linear Algebra Chapter 2 2.1 1. Not a linear transformation, since y 2 = x 2 + 2 is not linear in our sense. 3. Not linear, since y 2 = x 1 x 3 is nonlinear. 5. By Fact 2.1.2, the three columns of the 2 3 matrix A are T ( ~ e 1 ) , T ( ~ e 2 ), and T ( ~ e 3 ), so that A = 7 6 13 11 9 17 . 7. Note that x 1 ~v 1 + + x m ~v m = [ ~v 1 . . . ~v m ] x 1 x m , so that T is indeed linear, with matrix [ ~v 1 ~v 2 ~v m ]. 9. We have to attempt to solve the equation y 1 y 2 = 2 3 6 9 x 1 x 2 for x 1 and x 2 . Reducing the system 2 x 1 + 3 x 2 = y 1 6 x 1 + 9 x 2 = y 2 we obtain x 1 + 1 . 5 x 2 = 0 . 5 y 1 = 3 y 1 + y 2 . No unique solution ( x 1 , x 2 ) can be found for a given ( y 1 , y 2 ); the matrix is noninvertible. 11. We have to attempt to solve the equation y 1 y 2 = 1 2 3 9 x 1 x 2 for x 1 and x 2 . Reducing the system x 1 + 2 x 2 = y 1 3 x 1 + 9 x 2 = y 2 we find that x 1 = 3 y 1 2 3 y 2 x 2 = y 1 + 1 3 y 2 . The inverse matrix is 3 2 3 1 1 3 . 13. a. First suppose that a 6 = 0. We have to attempt to solve the equation y 1 y 2 = a b c d x 1 x 2 for x 1 and x 2 . ax 1 + bx 2 = y 1 cx 1 + dx 2 = y 2 a x 1 + b a x 2 = 1 a y 1 cx 1 + dx 2 = y 2 c ( I ) x 1 + b a x 2 = 1 a y 1 ( d bc a ) x 2 = c a y 1 + y 2 28 SSM: Linear Algebra Section 2.1 x 1 + b a x 2 = 1 a y 1 ( ad bc a ) x 2 = c a y 1 + y 2 We can solve this system for x 1 and x 2 if (and only if) ad bc 6 = 0, as claimed. If a = 0, then we have to consider the system bx 2 = y 1 cx 1 + dx 2 = y 2 swap : I II cx 1 + dx 2 = y 2 bx 2 = y 1 We can solve for x 1 and x 2 provided that both b and c are nonzero, that is if bc 6 = 0. Since a = 0, this means that ad bc 6 = 0, as claimed. b. First suppose that ad bc 6 = 0 and a 6 = 0. Let D = ad bc for simplicity. We continue our work in part (a): x 1 + b a x 2 = 1 a y 1 D a x 2 = c a y 1 + y 2 a D x 1 + b a x 2 = 1 a y 1 x 2 = c D y 1 + a D y 2 b a ( II ) x 1 = ( 1 a + bc aD ) y 1 b D y 2 x 2 = c D y 1 + a D y 2 x 1 = d D y 1 b D y 2 x 2 = c D y 1 + a D y 2 ( Note that 1 a + bc aD = D + bc aD = ad aD = d D . ) It follows that a b c d 1 = 1 ad bc d b c a , as claimed. If ad bc 6 = 0 and a = 0, then we have to solve the system cx 1 + dx 2 = y 2 bx 2 = y 1 c b x 1 + d c x 2 = 1 c y 2 x 2 = 1 b y 1 d c ( II ) x 1 = d bc y 1 + 1 c y 2 x 2 = 1 b y 1 It follows that a b c d 1 = d bc 1 c 1 b = 1 ad bc d b c a (recall that a = 0), as claimed. 29 Chapter 2 SSM: Linear Algebra 15. By Exercise 13a, the matrix a b b a is invertible if (and only if) a 2 + b 2 6 = 0, which is the case unless a = b = 0. If a b b a is invertible, then its inverse is 1 a 2 + b 2 a b b a , by Exercise 13b....
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This homework help was uploaded on 01/23/2008 for the course MATH 253B taught by Professor Bretsch during the Spring '08 term at Colby.
 Spring '08
 BRETSCH
 Linear Algebra, Algebra

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