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sh3lsh

  • one year ago

Determinant question

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  1. sh3lsh
    • one year ago
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    \[A =\left| \begin{array}{ccc} v_{1} \\ v_{2} \\ v_{3}\\v_{4}\end{array} \right| \det = 8\] What is the det of \[\left| \begin{array}{ccc} 6v_{1} +2v_{4} \\ v_{2} \\ v_{3}\\v_{4} + 3v_{1}\end{array} \right|\]

  2. terenzreignz
    • one year ago
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    Gotta love row operations ^^ Let's be general here: \[\det\left[ \begin{array}{ccc} v_{1} \\ v_{2} \\ v_{3}\\v_{4}\end{array} \right] =k\]

  3. terenzreignz
    • one year ago
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    Three elementary row operations: switch, multiply, and "add row multiple" All these have distinct effects on the determinant of the original matrix ^^

  4. sh3lsh
    • one year ago
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    Yeah! So I thought the answer would be that it would just be scaled by 3, because scalar addition or subtraction doesn't affect the determinant! However, the answer says it's 0, because the first row is twice the last. That didn't make much sense to me!

  5. terenzreignz
    • one year ago
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    When you switch one pair of rows, such as here: \[\det\left[ \begin{array}{ccc} v_{1} \\ \color{red}{v_{3}} \\ \color{blue}{v_{2}}\\v_{4}\end{array} \right] =-k\] the determinant gets negated. And will be negated for every pair of rows that you switch :D

  6. terenzreignz
    • one year ago
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    The first row is indeed twice the last. Are you finding it hard to believe that that causes the determinant to be zero? ^^

  7. sh3lsh
    • one year ago
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    Oops, scaled by 6! (v1 in first row is scaled by 6) Yeah, how does that change anything if the 4th row is a multiple of the first? Does that mean that the 4th row is redundant and if I rref'ed it the row would be 0, so if I tried to take the diagonals it would be 0?

  8. terenzreignz
    • one year ago
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    Okay. Let's take a weird approach to this. First, what I said: Switching a pair of rows causes the determinant to be negated or multiplied by a factor of (-1). Next, multiplying a row by a scalar in turn scales the determinant by that scalar. ie \[\det\left[ \begin{array}{ccc} v_{1} \\ \color{green}{n}v_{2} \\ v_{3}\\v_{4}\end{array} \right] =\color{green}{n}k\]

  9. terenzreignz
    • one year ago
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    So far, so good, yes? ^^

  10. terenzreignz
    • one year ago
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    So suppose we have a matrix where the first row is a multiple of the last row, and suppose its determinant was d. Like this \[\Large \det\left[ \begin{array}{ccc} mu_{3} \\ u_{1} \\ u_{2}\\u_{3}\end{array} \right]=h \]

  11. terenzreignz
    • one year ago
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    If we switch the first and fourth rows, as per the rule, the determinant gets multiplied by (-1) We get \[\Large \det\left[ \begin{array}{ccc} \color{red}{u_{3}} \\ u_{1} \\ u_{2}\\ \color{red}{mu_{3}}\end{array} \right]=\color{red}{-h}\]

  12. terenzreignz
    • one year ago
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    If we scale the first row by m, as per the other rule, the determinant gets scaled as well by a factor of m. \[\Large \det\left[ \begin{array}{ccc} \color{red}mu_{3} \\ u_{1} \\ u_{2}\\mu_{3}\end{array} \right]=-h\color{red}{(m)}\]

  13. terenzreignz
    • one year ago
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    Now we scale the fourth row by 1/m. Again, by the rules, this would also scale the determinant by a factor of 1/m. \[\Large \det\left[ \begin{array}{ccc} mu_{3} \\ u_{1} \\ u_{2}\\ \color{red}{\left(\frac1m\right)}mu_{3}\end{array} \right]=-h(m)\color{red}{\left(\frac1m\right)}\]

  14. terenzreignz
    • one year ago
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    Simplifying, we get \[\Large \det\left[ \begin{array}{ccc} mu_{3} \\ u_{1} \\ u_{2}\\u_{3}\end{array} \right]=-h\] We ended up with our original matrix, but THIS time, we have shown that its determinant is also equal to -h. That can only mean h = -h Or h = 0 Done ^^

  15. sh3lsh
    • one year ago
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    You are awesome! Thanks!

  16. terenzreignz
    • one year ago
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    No problem ^^

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