matrices

^ Video jockey.

now plzz reply any1!!

:P
sorry
:P
typin error

accha ab solve bhi kar lo yaar
:P

the ans is given simply 2.......
can u tell how did u do it??

i had a doubt
that the matrix \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}

will also satisfy the condn......

so in this case trace is 4...........how can it be only 2??

bole toh "2*n"

answer can't be simply two

''answer can't be simply two''

Lets see [6]:D

tr(A)= \sum_{k=1}^{n}a_{kk}= \sum_{k=1}^{n}(u_kv_k)

Let A^2 = [a"_{ij}]= \sum_{k=1}^{n}(a_{ik}a_{kj})= \sum_{k=1}^{n}(u_iv_ku_kv_j)= u_iv_j\sum_{k=1}^{n}v_ku_k= u_iv_j(tr(A))

Similarly A^5 = [a""_{ij}]= u_iv_j(tr(A))^4

But A^5 = 16A so tr(A)= 2,-2