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Let P = $$\left[ {\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \\ \end{matrix} } \right]$$ and Q = [qij] be two 3 $$ \times $$ 3 matrices such that Q – P5 = I3. Then $${{{q_{21}} + {q_{31}}} \over {{q_{32}}}}$$ is equal to :

JEE · Math · previous-year question

  1. A.15
  2. B.9
  3. C.135
  4. D.10correct

Answer

D. 10

Explanation

$$P = \left[ {\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \\ \end{matrix} } \right]$$ $${P^2} = \left[ {\begin{matrix} 1 & 0 & 0 \\ {3 + 3} & 1 & 0 \\ {9 + 9 + 9} & {3 + 3} & 1 \\ \end{matrix} } \right]$$ $${P^3} = \left[ {\begin{matrix} 1 & 0 & 0 \\ {3 + 3 + 3} & 1 & 0 \\ {6.9} & {3 + 3 + 3} & 1 \\ \end{matrix} } \right]$$ $${P^n} = \left[ {\begin{matrix} 1 & 0 & 0 \\ {3n} & 1 & 0 \\ {{{n\left( {n + 1} \right)} \over 2}{3^2}} & {3n} & 1 \\ \end{matrix} } \right]$$ $${P^5} = \left[ {\begin{matrix} 1 & 0 & 0 \\ {5.3} & 1 & 0 \\ {15.9} & {5.3} & 1 \\ \end{matrix} } \right]$$ $$Q = {P^5} + {{\rm I}_3}$$ $$Q = \left[ {\begin{matrix} 2 & 0 & 0 \\ {15} & 2 & 0 \\ {135} & {15} & 2 \\ \end{matrix} } \right]$$ $${{{q_{21}} + {q_{31}}} \over {{q_{32}}}} = {{15 + 135} \over {15}} = 10$$ Aliter $$P = \left( {\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} } \right) + \left( {\begin{matrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \\ \end{matrix} } \right)$$ $$P = {\rm I} + X$$ $$X = \left( {\begin{matrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \\ \end{matrix} } \right)$$ $${X^2} = \left( {\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \\ \end{matrix} } \right)$$ $${{X_3} = 0}$$ $${{P^5} = {\rm I} + 5X + 10{X^2}}$$ $${Q = {P^5} + {\rm I} = 2{\rm I} + 5X + 10{X^2}}$$ $$Q = \left( {\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ \end{matrix} } \right) + \left( {\begin{matrix} 0 & 0 & 0 \\ {15} & 0 & 0 \\ {15} & {15} & 0 \\ \end{matrix} } \right) + \left( {\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ {90} & 0 & 0 \\ \end{matrix} } \right)$$ $$ \Rightarrow \,\,Q = \left( {\begin{matrix} 2 & 0 & 0 \\ {15} & 2 & 0 \\ {135} & {15} & 2 \\ \end{matrix} } \right)$$

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