matrices

well no breakthroughs?????

for lower triangular matrix a_ij=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.

@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

use 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

|A| = w^3 - 3 + i = i - 2 ≠1??

Oh I didn't notice that one!