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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?

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    I know I've seen this exact question earlier this week... anyone see it? – 2017-01-20
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    work it out for $2\times 2$ matrix, $3\times 3,$ etc. See a pattern? – 2017-01-20
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    The biggest hint I can give is that $A$ has the effect of moving all entries of whatever matrix it multiplies towards the top-right corner. $A$ has the diagonal above the main diagonal full of ones. $A^2$ has the diagonal two above the main diagonal full of ones. $A^3$ has the diagonal three above the main diagonal full of ones, etc... – 2017-01-20

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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 $k=10$.

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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 $A^n=0$.