JEE Main 12 Jan 2019 Morning Question 23
Question: Let $ P= \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \\ \end{bmatrix} $ and $ Q=[q _{ij}] $ be two $ 3\times 3 $ matrices such that $ Q-P^{5}=I _3. $ Then $ \frac{q _{21}+q _{31}}{q _{32}} $ is equal to [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) 10
B) 135
C) 9
D) 15
Show Answer
Answer:
Correct Answer: A
Solution:
- Here, $ P= \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \\ \end{bmatrix} $ and $ Q=P^{5}+I _3 $
$ \therefore $ $ P^{2}= \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 6 & 1 & 0 \\ 27 & 6 & 1 \\ \end{bmatrix} $ $ P^{3}= \begin{bmatrix} 1 & 0 & 0 \\ 6 & 1 & 0 \\ 27 & 6 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 9 & 1 & 0 \\ 54 & 9 & 1 \\ \end{bmatrix} $
Similarly, $ P^{5}= \begin{bmatrix} 1 & 0 & 0 \\ 15 & 1 & 0 \\ 135 & 15 & 1 \\ \end{bmatrix} $
$ \therefore $ $ Q= \begin{bmatrix} 2 & 0 & 0 \\ 15 & 2 & 0 \\ 135 & 15 & 2 \\ \end{bmatrix} $
$ \Rightarrow $ $ \frac{q _{21}+q _{31}}{q _{32}}=\frac{15+135}{15}=10 $
