253Solutions3

Linear Algebra with Applications (3rd Edition)

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ma253 ,Fa l l 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 ,theoutputbecomes λy 1 λ 2 y 2 λy 3 ,wh ich is not λ times the original output. *4. Justdoit. .. 93 - 3 2 - 91 4 - 9 - 2 51 5 . 5. The images of the basis vectors are the columns of the matrix, so it is A = ± 76 - 13 11 9 17 ² . *6. Yes, it is linear, and the matrix is 14 25 36 . 9. Not invertible. I really enjoy being disappointed and outraged. – Dwight Macdonald
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*10. Invertible, and the inverse is ± 9 - 2 - 41 ² .(E a s yi fyouknowhow ,o f course!) *16. If we compute ± 30 03 ²± x 1 x 2 ² = ± 3x 1 3x 2 ² = 3 ± x 1 x 2 ² , we see that this scales every vector by 3 . The inverse must then be scaling by 1/3 ,soitis ± 1 3 0 0 1 3 ² . 18. Same deal; the inverse is scaling by 2 , i.e., ± 20 02 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 reFection on the x 2 = x 1 line; it’s its own inverse. 24–30. Moreofthesame. .. 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 ± 25 11 ² . ±inally, (c) asks for the inverse matrix, which, using the formula we now know, is just ± - 1 3 5 3 1 3 - 2 3 ² . The philosophical question confronting us today, class, is this: Is the stupidity of the media compulsory? – James Bowman
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Multiplying that by ± 144 51 ² gives ± 37 14 ² ,asexpected. *52. For part (a), we get ± 420 2100 ² , 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 ² , we see that the system is consistent only if b 2 = 5b 1 .Th i si sbecau sethe output vector needs to represent the same amount of money in the two currencies. For part (c), mathematically speaking there are in±nitely many solutions. If we allow negative amounts of Canadian dollars (resp., of Rand) to represent debts, then there are in±nitely 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 in±nite number of solutions.
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253Solutions3 - ma253, Fall 2007 - Problem Set 3 Solutions...

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