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- 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

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- anonymous

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- 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

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

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

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

- 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

+ - +
- + -
+ - +
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

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

3 * (-4) * |5 8|
|3 6|
and of course the determinant of that 2x2 is 5(6)-3(8)

- anonymous

wait what happened to.. (-1)^3+2?

- 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

top left corner is always a +, and it checkerboards from there

- anonymous

oh okay and then I just carry that 3 from the 1st expansion along the ride lol till the end.

- amistre64

yep, its stuck on there luck .... well something thats stuck lol

- amistre64

your reducing each matrix to a simpler form, the scalars from the previous attempts dont vanish

- anonymous

can you explain the "sign board" and little more clealy ? :) everything else makes sense

- 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

a 2x2 is the simplest case to use for an example; why do we have a subtraction sign in there?

- 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

ahhh gotcha :)

- amistre64

when the matrixes get large, the sign board is equivalent to using the (-1)^(n+m) adjustment :)

- anonymous

that's why that was used! lol that confused me, but it made sense.

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