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જો $P = \left[ {\begin{array}{*{20}{c}}
1&0&0 \\
3&1&0 \\
9&3&1
\end{array}} \right]$ અને $Q = [q_{ij}]$ એ $3\times3$ શ્રેણિક છે કે જેથી $Q -P^5 = I_3$. તો $\frac{{{q_{21}} + {q_{31}}}}{{{q_{32}}}} =$
$10$
$135$
$15$
$9$
Solution
${P^2} = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
6&1&0\\
{24}&6&1
\end{array}} \right]{P^3} = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
9&1&0\\
{54}&9&1
\end{array}} \right].\therefore {P^5} = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
{15}&1&0\\
{135}&{15}&1
\end{array}} \right]$
$Q = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
{15}&1&0\\
{135}&{15}&1
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
1&0&0\\
0&1&0\\
0&0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
2&0&0\\
{15}&2&0\\
{135}&{15}&2
\end{array}} \right]$
So $\frac{{{q_{21}} + {q_{31}}}}{{{q_{32}}}} = \frac{{15 + 135}}{{15}} = 10$