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यदि $P =\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1\end{array}\right]$ तथा $Q =\left[ q _{ ij }\right]$ दो ऐसे $3 \times 3$ आव्यूह हैं, कि $Q - P ^{5}= I _{3}$ है, तो $\frac{ q _{21}+ q _{31}}{ q _{32}}$ बराबर है
$10$
$135$
$15$
$9$
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 $\frac{{{q_{21}} + {q_{31}}}}{{{q_{32}}}} = \frac{{15 + 135}}{{15}} = 10$