P=(100310931) Q=(qij)3×3 Matrices such that Q−P5=I3 then q21+q31q32=............

P2=(100310931)(100310931)=(1006102761)P4=(1006102761)(1006102761)=(100121090121)P5=(100121090121)(100310931)=(1001510135151)Q−P5=I3⇒Q =I3+P5=(2001520135152)q21+q31q32=15+13515=10

P=(100310931) Q=(qij)3×3 Matrices such that Q−P5=I3 then q21+q31q32=............

P2=(100310931)(100310931)=(1006102761)P4=(1006102761)(1006102761)=(100121090121)P5=(100121090121)(100310931)=(1001510135151)Q−P5=I3⇒Q =I3+P5=(2001520135152)q21+q31q32=15+13515=10

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$P = (\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{matrix}) Q = {(q_{i j})}_{3 \times 3}$ Matrices such that $Q - P^{5} = I_{3} t h e n \frac{q_{21} + q_{31}}{q_{32}} = ............$

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$9$

$15$

$135$

$10$

answer is A.

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Detailed Solution

$P^{2} = (\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{matrix}) (\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{matrix}) = (\begin{matrix} 1 & 0 & 0 \\ 6 & 1 & 0 \\ 27 & 6 & 1 \end{matrix})$

$P^{4} = (\begin{matrix} 1 & 0 & 0 \\ 6 & 1 & 0 \\ 27 & 6 & 1 \end{matrix}) (\begin{matrix} 1 & 0 & 0 \\ 6 & 1 & 0 \\ 27 & 6 & 1 \end{matrix}) = (\begin{matrix} 1 & 0 & 0 \\ 12 & 1 & 0 \\ 90 & 12 & 1 \end{matrix})$

$P^{5} = (\begin{matrix} 1 & 0 & 0 \\ 12 & 1 & 0 \\ 90 & 12 & 1 \end{matrix}) (\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{matrix}) = (\begin{matrix} 1 & 0 & 0 \\ 15 & 1 & 0 \\ 135 & 15 & 1 \end{matrix})$

$Q - P^{5} = I_{3} \Rightarrow Q = I_{3} + P_{5} = (\begin{matrix} 2 & 0 & 0 \\ 15 & 2 & 0 \\ 135 & 15 & 2 \end{matrix}) \frac{q_{21} + q_{31}}{q_{32}} = \frac{15 + 135}{15} = 10$