1well no breakthroughs?????
let
l l
l 1 0 0 l
l l
A= l 3 w 0 l
l l
l 0 3-i w^2 l
l l
then
a) A^105 is a lower triangular matrix
b) A^106 is a upper triangular matrix
c) trace of A^109 is zero
d) trace of A^108 is w^108
w, w^2 are cube oots of unity
A is a 3X3 matrix
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18 Answers
21for lower triangular matrix aij=0 for for j>i
so @uttra clearly A^106 is a lower triangular matrix..how u r saying it as upper triagular?
and again identity matrix is both upper and lower triangular.
4may be u r right but I 've seen the definition of upper and lower triangular in TMH
it says only aij i < j is lower triangular
21@uttara-well i wud say check the definition of lower and upper triangular matrix.
and an identity matrix is both upper triangular and lower triagular matrix.
"A matrix which is simultaneously upper and lower triangular is diagonal. The identity matrix is the only matrix which is both upper and lower unitriangular"-- this is what wiki says.
So given answers r correct
4@ eraGON & utd4ever
I don't know where i made a mistake but somehow my ans doesn't match with urs
A3 = I
sO A105 = A108 = I
Identity matrix ≠lower triangular matrix
So A & D R wrong
A109 = A108 A = I A = A
Trace (A109 ) = 1 + W + W2 = 0
A106 = A105 A = I A = A --->upper triangular matrix
So ans shud be B & C
21use Cayley–Hamilton theorem
such that u getA^3=\begin{vmatrix} 1 & 0&0 \\ 0 & 1&0 \\ 0&0&1 \end{vmatrix}
Now A^{105}=\begin{vmatrix} 1 & 0&0 \\ 0 & 1&0 \\ 0&0&1 \end{vmatrix}
A^{106}=\begin{vmatrix} 1 & 0&0 \\ 3 & \omega &0 \\ 0&3-i&\omega^2 \end{vmatrix}
calculate A108 AND A109 accordingly
note-
1)identity matrix remains same no matter to wat power u raise it.
2)an identity matrix is both upper traigular and lower triagular matrix.
3)if u dunno know abt cayley hamilton theorem check this http://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem
*thats not a determinant btw...by mistake i used determinamt brackets instead of matrix one
1The determinant of a triangular matrix equals the product of the diagonal entries
1surely u must have made a calculation mistake
this is a triangular matrix lAl=1*w*w^2=1
1do u kno how to solve such ones
one thing common tat all have lAl=1....wonder will it help...[7]
39c) toh pakka as 1 + w + w^2 = 0(relation in complex numbers).
1 + \omega + \omega^2 = 1 + \frac{-1+\sqrt{3}i}{2} + \frac{-1-\sqrt{3}i}{2} = 1 - 1 = 0
(Trace is the sum of the elements on the main diagonal, we don't include secondary diagonal here)