ind the value of -6 -3 -5 5 7 -6 3 9 9

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ind the value of -6 -3 -5 5 7 -6 3 9 9

Mathematics
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determinant of this matrix?
-6
can you post a screenshot of the full thing?

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I can't im sorry
I'm confused what the full question is exactly
finding the value
it's only missing the brackets
ok so this? \[\Large \left|\begin{array}{ccc} -6 & -3 & -5\\ 5 & 7 & -6\\ 3 & 9 & 9\\ \end{array}\right|\]
yes
ok so this is a determinant
have you learned that term?
I don't think i have
|dw:1443314924015:dw|
ok here is one method to do a determinant of a 3x3 matrix
step 1) focus on the first two columns (circled) |dw:1443315068425:dw|
copy those columns to the right of the matrix bar |dw:1443315093066:dw| keep the order the same
we now have this extended matrix |dw:1443315131877:dw|
step 2) draw diagonals that start at the top and face south east. I circled the terms eg: -6, 7 and 9 are the first group of terms |dw:1443315174379:dw|
now multiply out the terms in each subgroup -6*7*9 = -378 -3*(-6)*3 = 54 -5*5*9 = -225
then add up the products -378+54+(-225) = -549
step 3) repeat the last step, but now we're going in the northeast direction |dw:1443315369919:dw|
multiply out the subgroups 3*7*(-5) = -105 9*(-6)*(-6) = 324 9*5*(-3) = -135 add up the products -105+324+(-135) = 84
the sum from step 2 was -549 the sum from step 3 was 84 subtract the sums like so -549-84 = -633 and that wraps up the problem
It's a long process, but over time it should come easier with more practice
thank you so much, it really seems a bit complicated but im starting to understand :3 @jim_thompson5910
this page has a similar example http://www.sparknotes.com/math/algebra2/systemsofthreeequations/section3.rhtml
Isn't that the same thing as doing the scalar triple product
volume of parallellpipedd
spaned by the three vectors
(A x B) * C
yes I think a 3x3 matrix determinant is involved somewhere in that
vector cross product scalar triple product determinant |dw:1443320363438:dw|
same thing, i just always remembered that way

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