Chapter 2 Notes - Chapter 2 ISM Linear Algebra Chapter 2...

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Chapter 2 ISM: Linear Algebra Chapter 2 2.1 1. Not a linear transformation, since y 2 = x 2 + 2 is not linear in our sense. 2. Linear, with matrix 0 2 0 0 0 3 1 0 0 3. Not linear, since y 2 = x 1 x 3 is nonlinear. 4. A = 9 3 - 3 2 - 9 1 4 - 9 - 2 5 1 5 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 . 6. Note that x 1 1 2 3 + x 2 4 5 6 = 1 4 2 5 3 6 x 1 x 2 , so that T is indeed linear, with matrix 1 4 2 5 3 6 . 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 ]. 8. Reducing the system x 1 + 7 x 2 = y 1 3 x 1 + 20 x 2 = y 2 , we obtain x 1 = - 20 y 1 + 7 y 2 x 2 = 3 y 1 - y 2 . 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 0 = - 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. 52
ISM: Linear Algebra Section 2.1 10. We have to attempt to solve the equation y 1 y 2 = 1 2 4 9 x 1 x 2 for x 1 and x 2 . Reducing the system x 1 + 2 x 2 = y 1 4 x 1 + 9 x 2 = y 2 we find that x 1 = 9 y 1 + 2 y 2 x 2 = - 4 y 1 + y 2 or x 1 x 2 = 9 - 2 - 4 1 y 1 y 2 . The inverse matrix is 9 - 2 - 4 1 . 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 . 12. Reducing the system x 1 + kx 2 = y 1 x 2 = y 2 we find that x 1 = y 1 - ky 2 x 2 = y 2 . The inverse matrix is 1 - k 0 1 . 13. a. First suppose that a = 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 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 = 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 = 0. Since a = 0, this means that ad - bc = 0, as claimed. 53
Chapter 2 ISM: Linear Algebra b. First suppose that ad - bc = 0 and a = 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 = 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 0 = 1 ad - bc d - b - c a (recall that a = 0), as claimed.

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