## monroe17 2 years ago 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

1. SithsAndGiggles

$\left|\begin{matrix}5&11&8&7\\ 3&-2&6&23\\ 0&0&0&-3\\ 0&4&0&17\end{matrix}\right|$ 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.

2. monroe17

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

3. 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

4. monroe17

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

5. 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

6. 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 :)

7. 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)

8. amistre64

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

9. monroe17

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

10. 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

11. amistre64

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

12. monroe17

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

13. amistre64

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

14. amistre64

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

15. monroe17

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

16. 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.

17. amistre64

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

18. 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

19. monroe17

ahhh gotcha :)

20. amistre64

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

21. monroe17

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