anonymous
  • anonymous
Need help with determinants of matrix. Question is in the attached file
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
So my question is can I just take the cofactor expansion of column 3. which will give me |dw:1444515136849:dw| right?
anonymous
  • anonymous
@amistre64

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
your pic is cut ...
anonymous
  • anonymous
the drawing?
anonymous
  • anonymous
at the end is just a question mark, its not part of the solution
amistre64
  • amistre64
i cant remember how to express the alternating (-1)^n in terms of the element picked from the ith row and jth column ...
amistre64
  • amistre64
+ - + - + - - + - + - + + - + - + - visually it is compiled like this
anonymous
  • anonymous
I dont know what that means
amistre64
  • amistre64
oh well, there was some 'summation' formula that is compact for it ... which element are we getting the cofactor expression for?
anonymous
  • anonymous
in the question it just says to get the determinant using cofactor expansion
anonymous
  • anonymous
so in the question I took the cofactor expansion of the 3rd column and I got what is in the drawing
amistre64
  • 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
amistre64
  • amistre64
3+3 that is ...
anonymous
  • 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?
anonymous
  • anonymous
|dw:1444516084912:dw|
anonymous
  • anonymous
but when I check my answer on determinant calculator its totally off
anonymous
  • anonymous
what am i doing wrong?
mathmate
  • mathmate
One good trick to remember, diagonal is always positive! Yes, 3 column is best!
amistre64
  • amistre64
\[...+(-1)^{(3+3)}~(-7)[... +(-1)^{(3+1)}~(1) \begin{pmatrix} -7&-10\\ -3&-8 \end{pmatrix} +... ]+...\]
amistre64
  • amistre64
you have one more cofactor to express
amistre64
  • amistre64
unless you simply stopped at this and decided to shortcut the 2x2 det
anonymous
  • anonymous
yes that what I tried to do
Empty
  • 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
amistre64
  • 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
anonymous
  • 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?
anonymous
  • anonymous
*sorry not "each" row, just one row or column
amistre64
  • 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 \[(-1)^{(2+1)}(c)(b)+(-1)^{(2+2)}(d)(a)=ad-bc\]
anonymous
  • anonymous
yes, one row or one column
mathmate
  • mathmate
|dw:1444517082554:dw| It's a personal preference, but I prefer to expand along the first column (because of the one's).
anonymous
  • anonymous
ok lets try that, let me do it on paper first
amistre64
  • amistre64
a 3x3 has a diagonalized shortcut similar to the 2x2 .. you just have to copy the first 2 columns
amistre64
  • amistre64
-(gfc+hga+ieb) a b c a b e f g e f g h i g h +(afi+bgg+ceh)
amistre64
  • amistre64
lol, forgot how to do the alphabet on that one ...
amistre64
  • amistre64
hij for the bottom row of course .. and then corrected for lack of any common sense
anonymous
  • anonymous
|dw:1444517589850:dw|
anonymous
  • anonymous
70(30+24)-(70+30)+(56-30)=3706
anonymous
  • anonymous
that is still now right, according to a determinant calculator. can anyone see what I did wrong?
anonymous
  • anonymous
well igtg for now, ill leave this question open for anyone else to figure it out.
mathmate
  • 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).
anonymous
  • anonymous
oh right, lol totally forgot about bedmas here. Thanks for the help!
mathmate
  • mathmate
You're welcome! :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.