I need help with linear algebra
we all do, its a shame how they present this stuff
UMMM question #18 part D
thats way to blurry for me to read; or i need bifocals ... 50/50 lol
lol hahaha ok i will resave it give me a sec
ok this one is better
yeah, much better; i feel young again :)
hahah good
id d(A,B) the determinate of AB ?
noo distance
im gonna have to look that up to be sure .... in the mean time, feel free to browse the gift shop :)
does this distance function have another name it goes by perchance?
hmmmm
i keep getting "distance matrix" that has nothing to do with this problem
finding distance btwn two vectors
how did you do it for #17?
ummm i didnt i only have to do #18
got it, i think http://tutorial.math.lamar.edu/Classes/LinAlg/InnerProductSpaces.aspx
d(A,B) = ||A-B||
A-B 1 -1 -1 1
ohhhh i cld have given u the formula
your keeping secrets from me arent you; make him sweat is your plan .... ok
but teh problem is the condition in the begginning
how do u work that out
the inner product? its pretty much self defined if you can read the notation
cuz i know the answer for #17 part D it 3sqrt(6)
\(a_{rc}\) tells us matrix A, row and column entry
amistre i gtg but i will be back to check the answer
Thanks :D
which in this case is 0 2*1.0 + 0.1 + 1.0 + 2*0.1 = 0
im gonna work and see if i can get that for #17 to see if im doing it right
got it ... maybe :) #17 -1 3 0 -2 4 -2 1 1 d(A,B) = ||A-B|| -1 5 3 -3 ^2 and add the results 1+25+9+9 = 54 ; now sqrt them. sqrt(54) = sqrt(9*6) = 3sqrt(6)
#18 would then be the same process ... A B 10 01 01 10 A-B = 1 -1 -1 1 ^2 that parts, add and sqrt 1+1+1+1 = 4; sqrt(4) = 2
Thanks Amistre. I was doing the same thing but somehow got my arithmetic wrong. Thanks again @amistre64
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