Let A=[[|gathered 1 0 0 3 1 0 9 3 1 gathered]]| and B=[b_ij],1 i,j 3.

Question

Let $A=[[|\begin{gathered} 1 \quad  0 \quad  0 \ 3 \quad  1 \quad  0 \ 9 \quad  3 \quad  1 \end{gathered}]]|$ and $B=[{b}_{ij}],1\le i,j\le 3$. If $B={A}^{99}-I$, then the value of $\frac{{b}_{31}-{b}_{21}}{{b}_{32}}$ is :

Accepted Answer

$\begin{gathered} A=P+I \ A=[[|\begin{array}{ccc}0 \quad  0 \quad  0 \ 3 \quad  0 \quad  0 \ 9 \quad  3 \quad  0\end{array}]]|+[[|\begin{array}{ccc}1 \quad  0 \quad  0 \ 0 \quad  1 \quad  0 \ 0 \quad  0 \quad  1\end{array}]]| \ A=P+I{P}^{3}=0 \ {A}^{99}=I+99P+{}^{99}{C}_{2}{P}^{2} \end{gathered}$

$\begin{gathered} {A}^{99}-I=99[[|\begin{array}{ccc}0 \quad  0 \quad  0 \ 3 \quad  0 \quad  0 \ 9 \quad  3 \quad  0\end{array}]]|+{}^{99}{C}_{2}[[|\begin{array}{ccc}0 \quad  0 \quad  0 \ 0 \quad  0 \quad  0 \ 9 \quad  0 \quad  0\end{array}]]| \ =[[|\begin{array}{ccc}0 \quad  0 \quad  0 \ 99	imes 3 \quad  0 \quad  0 \ 44550 \quad  99	imes 3 \quad  0\end{array}]]| \end{gathered}$

${b}_{31}=44550,{b}_{21}=99	imes 3,{b}_{32}=99	imes 3$

$	herefore $ $\frac{{b}_{31}-{b}_{21}}{{b}_{32}}$ = $\frac{44550-297}{297}=149$

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