3 and 4 .Determinants and Matrices
hard

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 $\frac{{{q_{21}} + {q_{31}}}}{{{q_{32}}}} =$

A

$10$

B

$135$

C

$15$

D

$9$

(JEE MAIN-2019)

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 $\frac{{{q_{21}} + {q_{31}}}}{{{q_{32}}}} = \frac{{15 + 135}}{{15}} = 10$

Standard 12
Mathematics

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