anonymous
  • anonymous
Sigma Notation Question: Is there any way to put this summation into Sigma Notation? I'm trying to make this a bit more succinct because I will eventually have to nest this to have 27 terms. \[\begin{vmatrix}a_{11} & 0 & 0 \\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33} \end{vmatrix} + \begin{vmatrix}0 & a_{12} & 0 \\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33} \end{vmatrix} + \begin{vmatrix}0 & 0 & a_{13} \\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33} \end{vmatrix}\]
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
imqwerty
  • imqwerty
so your gonna change only the 1st row like 1st u have a_{11} then a_{12} and u keep shifting them and u want summation till a_{27} right?
anonymous
  • anonymous
Basically, that's the first level of sums. So, take the first matrix there. I need to split that up even further like so: \[\begin{vmatrix}a_{11} & 0 & 0 \\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33} \end{vmatrix} = \begin{vmatrix}a_{11} & 0 & 0 \\a_{21} & 0 & 0\\a_{31} & a_{32} & a_{33} \end{vmatrix} + \begin{vmatrix}a_{11} & 0 & 0 \\0 & a_{22} & 0\\a_{31} & a_{32} & a_{33} \end{vmatrix} + \begin{vmatrix}a_{11} & 0 & 0 \\0 & 0 & a_{23}\\a_{31} & a_{32} & a_{33} \end{vmatrix}\]After I do that for the first 2 (I have 9 matrices at this point), for each of those 9, I need to do it again on the third row to get 27 matrices, so you can see why I want to try to use Sigma notation.
imqwerty
  • imqwerty
ok the summation is 0

Looking for something else?

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

More answers

imqwerty
  • imqwerty
note that evry determinant will have 2 zeroes in 2 diff columns whenever u see two 0s in two different columns then the determinant equal 0 can u find out y is it so :)
anonymous
  • anonymous
I'm not trying to calculate it though. I'm trying to write the proof of evaluating a 3x3 determinant using this method for class. Also, it wouldn't be 0 because there would still be one entry in every row, wouldn't there, which can row reduce to an upper triangular matrix?
anonymous
  • anonymous
Do you think it'd be fine to denote it as row vectors like this: Let \(v_1 = \begin{pmatrix}a_{11} & a_{12} & a_{13}\end{pmatrix}\), \(v_2 = \begin{pmatrix}a_{21} & a_{22} & a_{23}\end{pmatrix}\), \(v_3 = \begin{pmatrix}a_{31} & a_{32} & a_{33}\end{pmatrix}\) \[\begin{vmatrix}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33} \end{vmatrix} = \begin{vmatrix}v_1 \\ v_2 \\ v_3\end{vmatrix}\]And letting \(t_i\) represent the ith permutation matrix row, \[\begin{vmatrix}a_{11} & 0 & 0 \\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33} \end{vmatrix} = \begin{vmatrix}t_1 \\ v_2 \\ v_3\end{vmatrix}\]Would this work to make it into a sigma, or is there a cleaner way?

Looking for something else?

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