matrices help!
^ see attached
My thoughts were that if I can prove that R * R^t (its transpose) were equal to 1 then R^t = R^-1 (inverse)
R^T =[ 2/7 6/7 -3/7 ] [6/7 3/7 2/7] [3/7 2/7 6/7]
did you try to get the inverse?
wouldnt the first 6/7 on second row be neg?
I haven't tried to get the inverse yet.
yes negative im sorry my pc keeps hanging up on me lol
try to get the inverse
to get the inverse i have to find the determinant of the matrix and the transpose then i get a adjoint matrix then I divide the adjoint matrix by the determinant?
yes
k, brb have to try to do this on paper.
k
the answer would be the same as the R transpose
ffor the determinate R i got this
R^1= R adj / /R/
\[2/7(3/7*6/7-2/7*2/7)-6/7(-6/7*6/7-2/7*3/7)+-3/7(-6/7*2/7-3/7*3/7)\]
detR=/R/ = 1
is /R/ the adjoint of matrix? Sorry this stuff is pretty confusing to me
/R/ is the determinant of matrix R it is also det R=/A/
R adj= adjoint of R
okay. I tried to get the det R for the matrix above following http://analyzemath.com/Tutorial-System-Equations/determinants.html
which resulted in 1
like you said
ok thats good for the det R, now you can get the adjoint, try doing it
okay, brb, back to the note book haha
ok have fun lol
could you give me a little help, the books example on adjoint of the matrix is not very clear.
ok ill give you a good example hold on
okay, thanks again for the help
I think i may have gotten it, It came out the same as the tranpose matrix
if A=[ 1 2 3 ] [ 0 4 5 ] [ 1 0 6 ] you need to find the cofactor of each A11=[ 4 5 ] = 24 A12= -[ 0 5 ] [ 0 6] [ 1 6 ] =5 A13 =[ 0 4 ] [ 1 0 ] = -4 A21=-[ 2 3 ] [ 0 6 ]= -12 etc.... then you will have cofactor of A= [ 24 5 -4] [ -12 3 2] [-2 -5 4] then you can have adj A =[ 24 -12 -2 ] [ 5 3 -5] [ -4 2 4]
note here i use the co factor instead of the transpose
perfect I found a cofactor example too that help me along, thanks :)
yes the adj R is the same as transpose,,but they are not all the time equal....inthis case they are only equal
ok hope that help you out
perfect ! so i proved the second part! awesome thanks so much for your help
ok now have fun
yes, I guess my only other question is what did it mean by a proper rotation matrix?
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