a$)$ Show that the above equation can be written in the form of a matrix such that $(S,N)$ is

$S^{th}$ element of a $101$ matrix ${}_N$ :

$(N)=A(N-1)$

Once $(N)$ is calculated, $(S,N)$ can be found by reading $S^{th}$ element of $(N)$ matrix,

$(S,N)=(N{)}_S.$ For instance, $0^{th}$ element of $(2)$ is $(0,2)=1.$

Note that $(1)$ corresponds to the starting point before we move to the second digit of the num$-$

ber. So$,$ It is a $101$ matrix with all the elements equal to $1.$

$(1)=([1],[1],[1],[1],[1],[1],[1],[1],[1],[1])$

b$)$ Show that $(N)$ can be found recursively using the following equation:

$(N)=A$dots$AA(1)=A^{N-1}(1)$

That's it$!$ We solved the problem. Let's write a Python code to find $(N)$ for a given $N.$

c$)$ Create a $1010$ NumPy matrix which contains all the elements we have found for A$.$ Find

the transpose of the matrix using numpy.transpose. Are the matrix and its transpose equal?

Did you expect this result?

d$)$ Define a function with the name of "Dialer$\_$Problem" which takes two parameters, $S$ and $N$

and returns $(S,N).$ Within the function, use a "for loop" to multiply $(1)$ to A and then $A$

and so on$.$ Test your function for a few examples. For instance, running Dialer$\_$Problem $(5,10)$

should return $18713.$

e$)$ Rewrite the same code without using NumPy package. You will need to create a nested

list which includes all the elements in A and also define a function which performs matrix

multiplication.