anonymous
  • anonymous
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
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\left|\begin{matrix}5&11&8&7\\ 3&-2&6&23\\ 0&0&0&-3\\ 0&4&0&17\end{matrix}\right|\] 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.
anonymous
  • anonymous
it is new material ;/but i sorta understood what you just did.. but what do i do with that 3? enable to expand again?
amistre64
  • amistre64
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

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anonymous
  • anonymous
so starting off where @SithsAndGiggles left off.. would i do (3)(4)* [-1^{3+2}] of 5 11 3 -2
amistre64
  • amistre64
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
amistre64
  • amistre64
+ - + - + - + - + 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 :)
amistre64
  • amistre64
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)
amistre64
  • amistre64
3 * (-4) * |5 8| |3 6| and of course the determinant of that 2x2 is 5(6)-3(8)
anonymous
  • anonymous
wait what happened to.. (-1)^3+2?
amistre64
  • amistre64
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
amistre64
  • amistre64
top left corner is always a +, and it checkerboards from there
anonymous
  • anonymous
oh okay and then I just carry that 3 from the 1st expansion along the ride lol till the end.
amistre64
  • amistre64
yep, its stuck on there luck .... well something thats stuck lol
amistre64
  • amistre64
your reducing each matrix to a simpler form, the scalars from the previous attempts dont vanish
anonymous
  • anonymous
can you explain the "sign board" and little more clealy ? :) everything else makes sense
amistre64
  • amistre64
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.
amistre64
  • amistre64
a 2x2 is the simplest case to use for an example; why do we have a subtraction sign in there?
amistre64
  • amistre64
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
anonymous
  • anonymous
ahhh gotcha :)
amistre64
  • amistre64
when the matrixes get large, the sign board is equivalent to using the (-1)^(n+m) adjustment :)
anonymous
  • anonymous
that's why that was used! lol that confused me, but it made sense.

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