## Practice Final #1,2,3,4,5,7 with Pictures |
alexwang0518 |
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1) http://imgur.com/O0spyNr 2) http://imgur.com/ZTbc4tf 3) http://imgur.com/a/59lh0 4) http://imgur.com/dIHQqQt 5) http://imgur.com/RuwhhZe 7) http://imgur.com/K3YfL8C From reading through everyone else's posts, I think my three might not be the right answer. Also I did not make my #5 unit vectors, so my answer needs to be divided. |

Jackhe: June 2, 2015, 12:39 a.m. |
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In number 5b the given basis and vectors are in R^{3} right? Why did you use polynomials in the answer? |

CarlHarmon: June 2, 2015, 9:01 a.m. |
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He chose {$1,x,x^2$} as his basis for $R^3$. Which is correct, but there is an error in finding $v_2$ - when you divide by the length of $v_1$, you have to do the inner product. So instead of being 1, it's actually the integral of 1dx from 0 to 2; so 2. Then your $v_2$ should be $x-1$. |

alexwang0518: June 2, 2015, 1:21 p.m. |
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Yes, there is a mistake in my 5. The inner product defines length and perpendicularity so in my projection, the denominator needs to be the inner product give squared in the square root. Here is a correct write up I believe http://imgur.com/94fbrsh |

julianna_burke: June 2, 2015, 9:11 p.m. |
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I believe your mistake in #3 is when you are doing row reduction for E4 and in the second step you accidently drop the negative from your second 3. Fix the negative and you should get the same answer of those who have posted. I also believe you made a slight mistake in 7. At the very end, when you are doing the dot product between w2 and the vector you do -1*2=-1. |

dianejung_: June 3, 2015, 2:21 a.m. |
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For 3, it's just a little mistake. In the original problem it's a-3b+3c |

dianejung_: June 3, 2015, 2:27 a.m. |
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just kidding. I thought you were referring to something else. my bad |

jlumxajaha: June 3, 2015, 10:07 p.m. |
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I noticed the same thing as julianna_burke. You made a small mistake with the dot product. The final answer should be something like (59 122 69) or something close because I might have made an error too. |

alexwang0518: June 4, 2015, 1:38 p.m. |
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Yes, #3 has a row reduction in mistake. If the negative is added it will reduce to the correct eigenvector! |

omar77w: June 4, 2015, 4:21 p.m. |
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in 4b, I think you should take the absolute of the nominator. So it would be arccos = ($\frac{4}{9$\sqrt{2}$}$ |

aleksanderjanczewski: June 5, 2015, 12:57 a.m. |
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Problem 5 wrong, magnitude is calculated using definition for inner product |