## anonymous one year ago Need help with determinants of matrix. Question is in the attached file

1. anonymous

2. anonymous

So my question is can I just take the cofactor expansion of column 3. which will give me |dw:1444515136849:dw| right?

3. anonymous

@amistre64

4. amistre64

5. anonymous

the drawing?

6. anonymous

at the end is just a question mark, its not part of the solution

7. amistre64

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

+ - + - + - - + - + - + + - + - + - visually it is compiled like this

9. anonymous

I dont know what that means

10. amistre64

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

11. anonymous

in the question it just says to get the determinant using cofactor expansion

12. anonymous

so in the question I took the cofactor expansion of the 3rd column and I got what is in the drawing

13. amistre64

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

3+3 that is ...

15. anonymous

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

|dw:1444516084912:dw|

17. anonymous

but when I check my answer on determinant calculator its totally off

18. anonymous

what am i doing wrong?

19. mathmate

One good trick to remember, diagonal is always positive! Yes, 3 column is best!

20. amistre64

$...+(-1)^{(3+3)}~(-7)[... +(-1)^{(3+1)}~(1) \begin{pmatrix} -7&-10\\ -3&-8 \end{pmatrix} +... ]+...$

21. amistre64

you have one more cofactor to express

22. amistre64

unless you simply stopped at this and decided to shortcut the 2x2 det

23. anonymous

yes that what I tried to do

24. Empty

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

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

@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

*sorry not "each" row, just one row or column

28. amistre64

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

yes, one row or one column

30. mathmate

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

31. anonymous

ok lets try that, let me do it on paper first

32. amistre64

a 3x3 has a diagonalized shortcut similar to the 2x2 .. you just have to copy the first 2 columns

33. amistre64

-(gfc+hga+ieb) a b c a b e f g e f g h i g h +(afi+bgg+ceh)

34. amistre64

lol, forgot how to do the alphabet on that one ...

35. amistre64

hij for the bottom row of course .. and then corrected for lack of any common sense

36. anonymous

|dw:1444517589850:dw|

37. anonymous

70(30+24)-(70+30)+(56-30)=3706

38. anonymous

that is still now right, according to a determinant calculator. can anyone see what I did wrong?

39. anonymous

well igtg for now, ill leave this question open for anyone else to figure it out.

40. mathmate

@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

oh right, lol totally forgot about bedmas here. Thanks for the help!

42. mathmate

You're welcome! :)