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anonymous

  • one year ago

Need help with determinants of matrix. Question is in the attached file

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    So my question is can I just take the cofactor expansion of column 3. which will give me |dw:1444515136849:dw| right?

  3. anonymous
    • one year ago
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    @amistre64

  4. amistre64
    • one year ago
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    your pic is cut ...

  5. anonymous
    • one year ago
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    the drawing?

  6. anonymous
    • one year ago
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    at the end is just a question mark, its not part of the solution

  7. amistre64
    • one year ago
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    i cant remember how to express the alternating (-1)^n in terms of the element picked from the ith row and jth column ...

  8. amistre64
    • one year ago
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    + - + - + - - + - + - + + - + - + - visually it is compiled like this

  9. anonymous
    • one year ago
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    I dont know what that means

  10. amistre64
    • one year ago
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    oh well, there was some 'summation' formula that is compact for it ... which element are we getting the cofactor expression for?

  11. anonymous
    • one year ago
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    in the question it just says to get the determinant using cofactor expansion

  12. anonymous
    • one year ago
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    so in the question I took the cofactor expansion of the 3rd column and I got what is in the drawing

  13. amistre64
    • one year ago
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    http://www.wolframalpha.com/input/?i=cofactor+expansion it appears that you followed the formula correctly to me -1^(3+2) (-7) |..| yeah you did fine for that element

  14. amistre64
    • one year ago
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    3+3 that is ...

  15. anonymous
    • one year ago
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    So now I just have to get a cofactor expansion of any aij entry in the 3x3 matrix correct? so I took the a31 entry |dw:1444516001995:dw| is that right?

  16. anonymous
    • one year ago
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    |dw:1444516084912:dw|

  17. anonymous
    • one year ago
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    but when I check my answer on determinant calculator its totally off

  18. anonymous
    • one year ago
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    what am i doing wrong?

  19. mathmate
    • one year ago
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    One good trick to remember, diagonal is always positive! Yes, 3 column is best!

  20. amistre64
    • one year ago
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    \[...+(-1)^{(3+3)}~(-7)[... +(-1)^{(3+1)}~(1) \begin{pmatrix} -7&-10\\ -3&-8 \end{pmatrix} +... ]+...\]

  21. amistre64
    • one year ago
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    you have one more cofactor to express

  22. amistre64
    • one year ago
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    unless you simply stopped at this and decided to shortcut the 2x2 det

  23. anonymous
    • one year ago
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    yes that what I tried to do

  24. Empty
    • one year ago
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    Yeah you were able to throw away those other 3 determinants with your first step since you had 3 zeroes there, but now you don't have anymore zeroes so you have to fully evaluate the ones that are rest

  25. amistre64
    • one year ago
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    oh, i thought they were just randomly checking an element in the expansion .... didnt even consider that they were thinking this was the whole of it

  26. anonymous
    • one year ago
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    @empty so from when I have the 3x3 det in the second step. I have to get the cofactor expansion of each row or column? is that what you are trying to say?

  27. anonymous
    • one year ago
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    *sorry not "each" row, just one row or column

  28. amistre64
    • one year ago
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    we only need to go across one row right? \[\begin{pmatrix} a&b\\c&d \end{pmatrix}\] row 1 \[(-1)^{(1+1)}(a)(d)+(-1)^{(1+2)}(b)(c)=ad-bc\] or row2 \[(-1)^{(2+1)}(c)(b)+(-1)^{(2+2)}(d)(a)=ad-bc\]

  29. anonymous
    • one year ago
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    yes, one row or one column

  30. mathmate
    • one year ago
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    |dw:1444517082554:dw| It's a personal preference, but I prefer to expand along the first column (because of the one's).

  31. anonymous
    • one year ago
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    ok lets try that, let me do it on paper first

  32. amistre64
    • one year ago
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    a 3x3 has a diagonalized shortcut similar to the 2x2 .. you just have to copy the first 2 columns

  33. amistre64
    • one year ago
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    -(gfc+hga+ieb) a b c a b e f g e f g h i g h +(afi+bgg+ceh)

  34. amistre64
    • one year ago
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    lol, forgot how to do the alphabet on that one ...

  35. amistre64
    • one year ago
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    hij for the bottom row of course .. and then corrected for lack of any common sense

  36. anonymous
    • one year ago
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    |dw:1444517589850:dw|

  37. anonymous
    • one year ago
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    70(30+24)-(70+30)+(56-30)=3706

  38. anonymous
    • one year ago
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    that is still now right, according to a determinant calculator. can anyone see what I did wrong?

  39. anonymous
    • one year ago
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    well igtg for now, ill leave this question open for anyone else to figure it out.

  40. mathmate
    • one year ago
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    @m0j0jojo I suggest you multiply by -7 at the very end, then there will be less distribution to do. so the whole thing, instead of: 70(30+24)-(70+30)+(56-30) should read: -7(-10(30+24)-1(70+30)+1(56-30)) and you should get the correct answer (just a touch below 4300).

  41. anonymous
    • one year ago
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    oh right, lol totally forgot about bedmas here. Thanks for the help!

  42. mathmate
    • one year ago
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    You're welcome! :)

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