Question

$\text{Let}P=\left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]$ 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 $103$

$P=\left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]$
$\Rightarrow P^2=\left[\begin{array}{ccc}1& 0& 0\\ 4\times 2& 1& 0\\ 16(1+2)& 4\times 2& 1\end{array}\right]$

$\Rightarrow P^3=\left[\begin{array}{ccc}1& 0& 0\\ 4\times 3& 1& 0\\ 16(1+2+3)& 4\times 3& 1\end{array}\right]$

$⋮$
$⋮$
$\text{Similarly}{P}^{50}=\left[\begin{array}{ccc}1& 0& 0\\ 4\times 50& 1& 0\\ 16(1+2+3+...+50)& 4\times 50& 1\end{array}\right]$

$\Rightarrow P^{50}=\left[\begin{array}{ccc}1& 0& 0\\ 200& 1& 0\\ 20400& 200& 1\end{array}\right]$

As $P^{50}-Q=I,$
$\therefore \left[\begin{array}{ccc}1& 0& 0\\ 200& 1& 0\\ 20400& 200& 1\end{array}\right]-\left[\begin{array}{ccc}{q}_{11}& q_{12}& q_{13}\\ q_{21}& q_{22}& q_{23}\\ q_{31}& q_{32}& q_{33}\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$20400-q_{31}=0\Rightarrow q_{31}=20400$
$200-q_{32}=0\Rightarrow q_{32}=200$
$200-q_{21}=0\Rightarrow q_{21}=200$
$\therefore \frac{q_{31}+q_{32}}{q_{21}}=\frac{20400+200}{200}=103$