Solution 80: Slaying a Monstrous Determinant using Second Differences

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  • Published on Feb 21, 2019
  • We have a quadratic sequence embedded within the determinant -- Can we take advantage of this?
    This particular problem was inspired by one of the sections on determinant in "Linear Algebra Gems: Assets for Undergraduate Mathematics," a marvelous book I incidentally came upon in a university library, containing many fascinating articles on linear algebra. The article on determinant essentially used the techniques presented in the video, along with a slightly more sophisticated algebra, to derive the formula for a determinant of a square n x n matrix with its terms forming a sequence with (n-1)st common difference. I realized that the formula can be extended to cover the case where each term is written as a product of the aforementioned sequence and a geometric sequence, and the result was the problem in the video. If you are interested in exploring some "non-standard" yet fascinating theorems and ideas from linear algebra, I encourage you to check out the book from the nearest university library. =)
    Congratulations to Avi Uday, Benjamin Wang, aby p, Arun Bharadwaj, Andre Ben, Jaleb, 張惟淳, Serengeti Ghasa, Varun Shah, and mstmar for successfully solving this math challenge question! Avi Uday was the first person to solve the question.
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Comments • 19

  • LetsSolveMathProblems
    LetsSolveMathProblems  3 months ago +29

    * At 13:33, I should have written 6/80 = 3/40, not 1/15, for the final answer. I apologize for the error. Thank you, Hiren Bavaskar, for notifying me!

  • JT
    JT 2 months ago

    In solving this problem you took at least 3 side-steps to solve the problem...and these are all excellent and necessary. When I pointed out in an earlier video you became very angry. saying that you don;t explain every point. Well you just did....e.x..p.l..a..i...n e..v..e..r..y. p..o..i..n....t.

  • Monster Azice
    Monster Azice 3 months ago +1

    Where does he get these questions? They are super interesting

    • LetsSolveMathProblems
      LetsSolveMathProblems  3 months ago +2

      This particular problem was inspired by one of the sections on determinant in "Linear Algebra Gems: Assets for Undergraduate Mathematics," a marvelous book I incidentally came upon in a university library, containing many fascinating articles on linear algebra. The article on determinant essentially used the techniques presented in the video, along with a slightly more sophisticated algebra, to derive the formula for a determinant of a square n x n matrix with its terms forming a sequence with (n-1)st common difference. I realized that the formula can be extended to cover the case where each term is written as a product of the aforementioned sequence and a geometric sequence, and the result was the problem in the video.
      I have authored most of the Weekly Math Challenges so far--The exceptions owe their existence to my younger brother or both of us working in conjunction. I'm glad you find them interesting! =)

  • Louis Thirion
    Louis Thirion 3 months ago

    You cannot take the cube root or some other root to reduce a fraktion...
    Therefore 216/80^3=6^3/80^3 is not equivalent to 6/80.
    The correct answer would be 27/6400.

    • Sanjay Kannan
      Sanjay Kannan 3 months ago +2

      The question is to find the cube root of the determinant.

  • turtlellamacow
    turtlellamacow 3 months ago

    Nice problem and very nice explanation as always. I think it would have been a bit cleaner to simply pull all the r's out of the matrix at the beginning (for an overall r^15 outside) rather than carry them along through all the manipulations.

  • adandap
    adandap 3 months ago +2

    TBH, in this case I think that simply calculating the determinant first and then using a little algebra wasn't any more difficult than the (admittedly more elegant) solution presented here.
    The independence from a is undoubtedly the underlying reason for the very nice cancellations I found - but is there any way to see that from the initial form of the matrix?

    • Vaidhyanathan A
      Vaidhyanathan A 3 months ago

      @LetsSolveMathProblems how is it to author such beautiful problems,it's really marvellous,answering difficult questions-great,but creating difficult questions is much beyond, 👌👌👌

    • adandap
      adandap 3 months ago

      Thank you for your response @LetsSolveMathProblems. It begs the obvious generalisation to higher powers - how long will the neat progression save us from an a dependence, or is the step up to cubic enough? I put it into Mathematica to see - the answer is 'no'. Explicitly you get r^15 * 972 (24 + 38 a + 12 a^2 + a^3). Following your reasoning, I thought that a 3x3 with a cubic kind of exhausts that ability of the progression to 'help'. So I generalised it to a 4x4 matrix with cubic terms a^3, (a+1)^3 ... (a+15)^3 etc - and, sure enough, it's independent of a when you do that. But a quartic progression has a determinant which is quartic in a, etc.

    • LetsSolveMathProblems
      LetsSolveMathProblems  3 months ago

      If the terms of the matrix formed an arithmetico-geometric sequence (instead of a quadratic-geometric sequence, as in the featured problem), it may be intuitively seen that the determinant must be 0 because successive column subtractions will soon yield a 0 column. Perhaps it could be expected that increasing the degree by 1 (from linear to quadratic) would not be quite strong enough to disrupt the determinant's independence from a.

  • alex zorba
    alex zorba 3 months ago +1

    superb!

  • Angry Tomato!
    Angry Tomato! 3 months ago +10

    This is so me at exams, do all hard work good, screw up on easy stuff on end...

  • Quantum Tlisen
    Quantum Tlisen 3 months ago +3

    A matrix! That's really nice change of pace in my opinion. Nice video!

  • shashvat singh
    shashvat singh 3 months ago +3

    I am in class 10 and i can understand this because the explanation was very nice!

  • alain Rogez
    alain Rogez 3 months ago +1

    I tried on my own by hand but it was by far too hard for me.

  • Hiren Bavaskar
    Hiren Bavaskar 3 months ago +26

    6/80 is not 1/15.. rather it is 3/40 ..please correct it

    • Hiren Bavaskar
      Hiren Bavaskar 3 months ago +1

      @LetsSolveMathProblems No problem :) Btw I like ur problems.. Keep up the good work!

    • LetsSolveMathProblems
      LetsSolveMathProblems  3 months ago +4

      Oops! Thank you so much for notifying me! I can't believe I made an arithmetic mistake right at the end. I just posted a pinned comment with the correction. =)