Imagine we have started from number $S,$ so $N-1$ digits are left to be dialed. Starting from $S,$

we can only dial an allowed set of numbers, $\{j|A_{Sj}=1\}.$ Therefore, $(S,N)$ can be calculated

recursively by adding $(k,N-1)$ values for kin$\{j|A_{Sj}=1\}.$

$(S,N)={\displaystyle \underset{\text{k}\text{i}\text{n}\{j|A_{Sj}=1\}}{\overset{?}{}}}(k,N-1)$

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