## practice midterm #2 |
wllmskn |
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I'm having trouble reducing the matrix in this problem. I can't get the second column into reduced row echelon form without introducing variables back into the bottom of the first column. Did anyone get this? |

wllmskn: May 3, 2015, 2:09 p.m. |
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Are we allowed to switch rows when reducing a matrix that represents a linear transformation? That would solve the problem, but it seems like that would change its identity. |

Noeland: May 3, 2015, 2:24 p.m. |
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We definitely can switch rows when doing the reduce. It is one of the three elementary row operations, which are 1) multiply a row by a non-zero constant. 2) replacement of rth row by rth row plus a scalar times sth row. 3) switch between two rows, and those two rows even don't need to be adjacent. |

wllmskn: May 3, 2015, 4:06 p.m. |
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Ok thanks! Did you get (1 0 0 -7/6) (0 1 0 17/3) (0 0 1 -1/2) for the echelon form? Just to check if I did it right.. |

valerie: May 3, 2015, 6:13 p.m. |
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I also got (1 0 0 -7/6) (0 1 0 17/3) (0 0 1 -1/2) for the echelon form. |

wllmskn: May 3, 2015, 7:31 p.m. |
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Great, thanks! And then we get the dimension of the image from the rank, and use the rank nullity theorem to get the rank of the kernel? do we have to do anything else to prove why this works? |

carlintou: May 3, 2015, 8:20 p.m. |
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Nope that should be all. So the dim(im(A)) = 3 and the dim(ker(A)) = 1 |

Jackhe: May 4, 2015, 12:56 p.m. |
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a simple way to check your rref is to plug it in into your TI (if you have one, I have TI-84 plus) and let it compute the rref. |

brookewenig: May 4, 2015, 10:02 p.m. |
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I used this website to double check my RREF for my calculations: http://www.math.purdue.edu/~dvb/matrix.html. It can solve up to 5x6 matrices. |

danielkim: May 4, 2015, 10:39 p.m. |
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You can easily use the rank nullity theorem to obtain dim(ker(A)). Remember the formula (# of columns in a matrix) = dim(ker(A)) + dim(im(A)). A is a 3x4 matrix, so it has 4 columns. Once you get the dimension of image of A, which is 3, you can plug it in the equation. 4 = 3 + dim(ker(A)) so 1 = dim(ker(A)) |

yulduzkhonbruin: May 5, 2015, 6:24 p.m. |
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I have the same answer for rref, but we also have to equate the last variable, let's call it w=t (t is a free variable), so the answer is the same, x=7t/6, y=-17t/3 and z=t/2. |

emmyralds: May 5, 2015, 9:17 p.m. |
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When you guys are completing the last part, "Find all vectors that map to zero when multiplied by A," we are essentially finding the ker(A) right? How did you guys do that? |