problem related to matrix 
I tried up to $3\times{3}$ matrix there it satisfied by order of $2$ but if in cases like this $10\times{10}$ or $20\times{20}$ how will we find out the order?
 A: Consider, for any integer $n\ge2$, the matrix :
$$A=[\delta_{i,j+1}]_{1\le i,j\le n}$$
whose entries are all zero except the ones just above the diagonal.
Note that $\delta_{a,b}$ is the usual Kronecker's symbol, defined by :
$$\delta_{a,b}=\cases{1\quad\mathrm{if}\,a=b\cr 0\quad\mathrm{otherwise}}$$
It can be proved by induction that, for each integer $p$ such that $1\le p\le n$, we have :
$$A^p=[\delta_{i,j+p}]$$
and, in particular, $A^n=0$ but $A^{n-1}\neq 0$.
You're asking what happens in the special case $n=10$ ... and so the correct answer is $k=10$.
A: Let $e_1,e_2,...e_n$ be the canonical basis of $\mathbb{R^n}$ ($e_k$ being a vector of zeroes except at position $k$ where there is a $1$.)
As we know that the columns of a matrix are the images of this canonical basis, we have:
$Ae_1=0,Ae_2=e_1,Ae_3=e_2,...Ae_n=e_{n-1}.$
Now left-multiply all these equations by $A$; you get:
$A^2e_1=0, A^2e_2=Ae_1=0, A^2e_3=Ae_2=e_1, ...A^2e_n=Ae_{n-1}=e_{n-2}.$
Then again multiply by $A$ the previous set of equations, etc.
At each new multiplication, the number of zeros increases by one.
Thus, if you do it $n$ times, you will end up with:
$A^ne_1=0, A^ne_2=0, A^ne_3=0, ...A^ne_n=0.$
which means that $A^n=0$.
