253Solutions3

# Linear Algebra with Applications (3rd Edition)

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ma253 , Fall 2007 — Problem Set 3 Solutions 1. Problems from the textbook: a. Section 2 . 1 1. Not linear. For example, 0 0 0 doesn’t go to 0 0 0 . 3. Not linear. If we scale x 1 x 2 x 3 by λ , the output becomes λy 1 λ 2 y 2 λy 3 , which is not λ times the original output. *4. Just do it. . . 9 3 - 3 2 - 9 1 4 - 9 - 2 5 1 5 . 5. The images of the basis vectors are the columns of the matrix, so it is A = 7 6 - 13 11 9 17 . *6. Yes, it is linear, and the matrix is 1 4 2 5 3 6 . 9. Not invertible. I really enjoy being disappointed and outraged. – Dwight Macdonald

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*10. Invertible, and the inverse is 9 - 2 - 4 1 . (Easy if you know how, of course!) *16. If we compute 3 0 0 3 x 1 x 2 = 3x 1 3x 2 = 3 x 1 x 2 , we see that this scales every vector by 3 . The inverse must then be scaling by 1/3 , so it is 1 3 0 0 1 3 . 18. Same deal; the inverse is scaling by 2 , i.e., 2 0 0 2 . 19. This is the projection onto the x -axis, hence not invertible. *20. This swaps x 1 and x 2 coordinates, so it’s the reflection on the x 2 = x 1 line; it’s its own inverse. 24–30. More of the same. . . 40. They are all given by multiplication by 1 × 1 matrices, i.e., by numbers. So they are all T ( x ) = ax for some a . Their graphs are lines through the origin. *49. Part (a) is an easy linear system; the answer is 37 and 14 . Part (b) asks for the transformation; it is the one given by the matrix 2 5 1 1 . Finally, (c) asks for the inverse matrix, which, using the formula we now know, is just - 1 3 5 3 1 3 - 2 3 . The philosophical question confronting us today, class, is this: Is the stupidity of the media compulsory? – James Bowman
Multiplying that by 144 51 gives 37 14 , as expected. *52. For part (a), we get 420 2100 , which is the total value of our money, given in C\$ and in R. It’s easy to see that the matrix is not invertible, for example because its determinant is 0 . If we row-reduce with an arbitrary right hand row b 1 b 2 , we see that the system is consistent only if b 2 = 5b 1 . This is because the output vector needs to represent the same amount of money in the two currencies. For part (c), mathematically speaking there are infinitely many solutions. If we allow negative amounts of Canadian dollars (resp., of Rand) to represent debts, then there are infinitely many solutions that make sense in the real world. If we insist on positive quantities, then because money can’t be fractioned into less than hundredths, there are many but not an infinite number of solutions.

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