28 thoughts on “Solving Ax=b | MIT 18.06SC Linear Algebra, Fall 2011

  • January 25, 2012 at 3:16 pm
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    Thanks for posting and i love watching this but sad to say that i am starting to lose you around 3.48… hahaha sorry… need to rewatch few more time i guess.

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  • February 3, 2012 at 12:04 am
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    @HubertSvab91 It is definitely South Slavic. Czech would have slightly different accent.

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  • February 27, 2013 at 2:06 am
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    there's an error, when she eliminates the last row, in fact we're adding the 2nd row to the third, but in the b's she subtracts the 2nd row from the third.
    So the correct third row is [ 0 0 0 | -6b1 + b2 + b3 ] so that the condition for the compatibility is (-6b1 + b2 + b3)=0

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  • March 7, 2013 at 9:37 pm
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    I don't understand what do you mean

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  • March 13, 2013 at 1:32 pm
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    She is Croatian, not Serbian.

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  • June 15, 2013 at 11:37 am
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    I lke the of man thanks you

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  • July 20, 2014 at 6:13 pm
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    with all this eastern european sensation, MIT must be really a good school

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  • February 16, 2015 at 5:19 pm
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    @JNRTomcat Nope, she was correct. We didn't actually add row 2 to row 3. If we did as you suggest, we would get -1 + -1 = -2. So our new row 3 would read [0 -2 0] and that's not what we want; we're trying to zero out the entire 3rd row. Instead, we have to SUBTRACT row 2 from row 3 which gets us -1 – (-1); the negatives cancel out and we get -1 + 1 = 0; that's exactly what we want. Therefore, we perform the same operation on our augmented column which gives us their difference, NOT their sum. (-4b1 + b3) – (-2b1 + b2) = -2b1 – b2 + b3.

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  • October 1, 2015 at 6:04 pm
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    so in real life you were asking whether Ax=b has a solution for all choice of b, the answer is no solution of Ax=b, because at the very end the conclusion is 0 = 2b1 -b2 + b3 <- not a solution

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  • January 12, 2016 at 3:31 am
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    very well explained, thanks a lot!

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  • January 12, 2016 at 9:18 pm
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    Is the special solution wrong here? If z=1 then we still have y=2b_1-b_2, or have I got it wrong?

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  • September 2, 2016 at 5:11 pm
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    Definitely not something to watch. I was picking it up at first but then I got all confused because she did.

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  • November 4, 2016 at 3:29 am
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    linear algebra with rapunzel 🙂

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  • December 14, 2016 at 5:17 am
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    is it mandatory to assume z=0

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  • January 1, 2017 at 8:30 am
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    I guess I'm the DESIGNATED Superficial Commenter here.. so .. here goes… I had to watch this VIDEO twice.. because the First time through I was totally Taken by Martina's Sex Appeal… her features and her cuteness… (I'm such a guy!!) .. anyway.. the second time through I did follow the lecture and I did Learn!!.. so.. BRAVO.. I like how Martina says MOOtiply for Multiply.. lol…. I wonder if Chekov was the camera guy …

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  • March 23, 2017 at 11:16 am
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    You should explain the particular solution more in depth.

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  • April 19, 2017 at 4:07 am
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    Las matemáticas son el lenguaje universal:DDDDDDD

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  • May 23, 2017 at 1:17 pm
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    can you help in solving a problem of FEA ,

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  • September 13, 2017 at 3:29 am
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    I like this video so much!

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  • November 16, 2017 at 8:25 pm
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    Garbage explanation. Why not take your time trying to explain EXACTLY why things happen instead of just trying to look smart? Example: WHY is this the case for the particular solution?

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  • March 1, 2018 at 11:35 pm
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    Great job, but you lost me on the special solution.

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  • May 18, 2018 at 1:22 am
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    is there and error at minute 5:22 ? I think she multiplies -2 (-2b1+b2) instead of -2(2b1-b2)

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  • September 2, 2018 at 10:24 pm
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    whut?

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  • September 3, 2018 at 1:45 pm
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    she seems over-educated

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  • September 25, 2018 at 8:57 am
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    Very clear , thanks!

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  • February 13, 2019 at 3:31 pm
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    beauty and brain thank you!

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  • February 19, 2019 at 12:30 pm
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    How about a parametric solution?

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  • October 5, 2019 at 8:52 pm
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    why in the special solution vector was it 2 and not -2 ?

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