Question

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 $Q=\left[q_{ij}\right]$ is a matrix, such that $p^{50}-Q=I$, then $\frac{q_{31}+q_{32}}{q_{21}}$ equals

• A. $52$
• B. $103$
• C. $201$
• D. $205$

Answer 1

The correct option is B $103$

$P=\left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]$

$P^2=\left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]$$\left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]$$=\left[\begin{array}{ccc}1& 0& 0\\ 8& 1& 0\\ 48& 8& 1\end{array}\right]$

$P^3=\left[\begin{array}{ccc}1& 0& 0\\ 8& 1& 0\\ 48& 8& 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\\ 12& 1& 0\\ 56& 12& 1\end{array}\right]$

$P^4=\left[\begin{array}{ccc}1& 0& 0\\ 8& 1& 0\\ 48& 8& 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\\ 16& 1& 0\\ 160& 16& 1\end{array}\right]$

$\therefore P^{50}=\left[\begin{array}{ccc}1& 0& 0\\ 4\times 50& 1& 0\\ p_{31}& 4\times 50& 1\end{array}\right]$$Q=\left[\begin{array}{ccc}{q}_{11}& q_{12}& q_{13}\\ q_{21}& q_{22}& q_{23}\\ q_{31}& q_{32}& q_{33}\end{array}\right]$

$P^{50}-Q=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\Rightarrow 4\times 50-q_{21}=0$

$p_{31}-q_{31}=0$

$4\times 50-q_{32}=q$

On observing,

$p_{{31}^1}=16\times 1$

$p_{{31}^2}=16\times (1+2)$

$p_{{31}^3}=16\times (1+2+3)$

$p_{{31}^{50}}=16\times (1+2+3+.....50)$

$=16\times \frac{50\times 51}{2}$

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

$=\frac{16\times 25\times 51+4\times 50}{4\times 50}$

$=103$