Matrices And Determinants Ques 28
Let $P=\left[\begin{array}{ccc}1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1\end{array}\right]$ and $I$ be the identity matrix of order 3 . If $Q=\left[q_{i j}\right]$ is a matrix, such that $P^{50}-Q=I$, then $\frac{q_{31}+q_{32}}{q_{21}}$ equals
(2016 Adv.)
(a) $52$
(b) $103$
(c) $201$
(d) $205$
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Answer:
Correct Answer: 28.(b)
Solution:
Formula:
Properties of matrix multiplication:
- Here,
$P=\left[\begin{array}{ccc}1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1\end{array}\right]$
$ \begin{alignedat} \therefore \quad P^2 & =\left[\begin{array}{ccc} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 [paste corrected text here] 1 & 0 & 0 \\ 4+4 & 1 & 0 \\ 16+32 & 4+4 & 1 \end{array}\right] \\ & =\left[\begin{array}{ccc} 1 & 0 & 0 \\ 4 \times 2 & 1 & 0 \ 16(1+2) & 4 \times 2 & 10 \end{array}\right] \\ \text { and } P^3 & =\left[\begin{array}{ccc} 1 & 0 & 0 \\ 4 \times 2 & 1 & 0 \ 16(1+2) & 4 \times 2 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{array}\right] \\ & =\left[\begin{array}{ccc} 1 & 0 & 0 \\ 4 \times 3 & 1 & 0 \ 16(1+2+3) & 4 \times 3 & 1 \end{array}\right] \end{aligned} $
From symmetry,
$ \begin{aligned} & P^{50}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 4 \times 50 & 1 & 0 \\ 16(1+2+3+\ldots+50) & 4 \times 50 & 1 \end{array}\right] \\ & \because \quad P^{50}-Q=I \\ Given & \therefore\left[\begin{array}{ccc}28-q_{11} & -q_{12} & -q_{13} \28-q_{21} & 1-q_{22} & -q_{23} \ 16 \times \frac{50}{2}(51)-q_{31} & 200-q_{32} & 1-q_{33} \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \Rightarrow \quad 200-q_{21}=0, \frac{16 \times 50 \times 51}{2}-q_{31}=0 \text{, } \\ & 200 - q_{32} = 0 \\ & \therefore \quad q_{21}=200, q_{32}=200, q_{31}=20400 \\ & \end{aligned} $
$\text { Thus, } \quad \frac{q_{31}+q_{32}}{q_{21}}=\frac{20400+200}{200}=\frac{20600}{200}=103$
BETA
