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monroe17

use cofactor expansion to solve for the determinant of the matrix 5 11 8 7 3 -2 6 23 0 0 0 -3 0 4 0 17

  • one year ago
  • one year ago

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  1. SithsAndGiggles
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    \[\left|\begin{matrix}5&11&8&7\\ 3&-2&6&23\\ 0&0&0&-3\\ 0&4&0&17\end{matrix}\right|\] Using a cofactor expansion will be significantly simpler if you use the third row. So, the expansion is \[\left|\begin{matrix}5&11&8&7\\ 3&-2&6&23\\ \color{red}0&\color{red}0&\color{red}0&\color{red}{-3}\\ 0&4&0&17\end{matrix}\right|=0C_{3,1}+0C_{3,2}+0C_{3,3}+(-3)C_{3,4}\] where C_{i,j} is the cofactor of a_{i,j}, and the cofactor is \[C_{i,j}=(-1)^{i+j}M_{i,j},\] where M_{i,j} is the minor of a_{i,j}. I hope none of this is new material. Right away, you can simplify the expansion to \[\begin{align*}\left|\begin{matrix}5&11&8&7\\ 3&-2&6&23\\ \color{red}0&\color{red}0&\color{red}0&\color{red}{-3}\\ 0&4&0&17\end{matrix}\right|&=(-3)(-1)^{3+4}\left|\begin{matrix}5&11&8\\3&-2&6\\0&4&0\end{matrix}\right|\\ &=3\left|\begin{matrix}5&11&8\\3&-2&6\\0&4&0\end{matrix}\right| \end{align*}\] Do you think you can take it from here? I'd suggest another expansion using the third row.

    • one year ago
  2. monroe17
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    it is new material ;/but i sorta understood what you just did.. but what do i do with that 3? enable to expand again?

    • one year ago
  3. amistre64
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    the 4x4 was effectively reduced to a 3x3, which is much simpler to play with. you can use that last row that is full of zeros to reduce it again

    • one year ago
  4. monroe17
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    so starting off where @SithsAndGiggles left off.. would i do (3)(4)* [-1^{3+2}] of 5 11 3 -2

    • one year ago
  5. amistre64
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    the matrix we have to work with is: 5 11 8 3 -2 6 0 4 0 lets cross out the row and colum with the 4 in it 5 - 8 3 - 6 - - - whats left is the matrix we want to attach to the -4 part; but there is a +-+- to sort out as well

    • one year ago
  6. amistre64
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    + - + - + - + - + notice that this sign board has a - in the place where the -4 would be; that tells us that we want -(-4) as the scalar so it looks like you had it worked our correctly other than the submatrix parts :)

    • one year ago
  7. amistre64
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    why do i see a -4 in there? now that i reread it lol, the sign board has a - where the 4 is; giving us a scalar of -(4)

    • one year ago
  8. amistre64
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    3 * (-4) * |5 8| |3 6| and of course the determinant of that 2x2 is 5(6)-3(8)

    • one year ago
  9. monroe17
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    wait what happened to.. (-1)^3+2?

    • one year ago
  10. amistre64
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    thats another way to represent the sign board; I never remember the power stuff so I just work out it out with the sign board

    • one year ago
  11. amistre64
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    top left corner is always a +, and it checkerboards from there

    • one year ago
  12. monroe17
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    oh okay and then I just carry that 3 from the 1st expansion along the ride lol till the end.

    • one year ago
  13. amistre64
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    yep, its stuck on there luck .... well something thats stuck lol

    • one year ago
  14. amistre64
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    your reducing each matrix to a simpler form, the scalars from the previous attempts dont vanish

    • one year ago
  15. monroe17
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    can you explain the "sign board" and little more clealy ? :) everything else makes sense

    • one year ago
  16. amistre64
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    the sign board is a term im calling it for lack of a better term :) the manner in which a determinant is worked out ... culling the parts from the matrix ....results in either adding or subtracting certain amounts. the sign board is a visual account of not shuffling the pieces around really.

    • one year ago
  17. amistre64
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    a 2x2 is the simplest case to use for an example; why do we have a subtraction sign in there?

    • one year ago
  18. amistre64
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    take the example: 0 4 3 0 the determinant of this using submatrix parts is; lets use the 4 and cancel out the row and column its in - - 3- this gives us 4(3) as the determinant if we dont adjust for signs but we know that the actual determinant is 0(0) - 4(3) = -12 the mechanics of the matrix present us with this sign board such that for every scalar we want to use we have to adjust for its placement: + - - + we used the scalar for the placement if the 4, and its submatrix (3) giving us a -4(3) determinant

    • one year ago
  19. monroe17
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    ahhh gotcha :)

    • one year ago
  20. amistre64
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    when the matrixes get large, the sign board is equivalent to using the (-1)^(n+m) adjustment :)

    • one year ago
  21. monroe17
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    that's why that was used! lol that confused me, but it made sense.

    • one year ago
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