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

Here, $P=\left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]$
$\therefore P^2=\left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 4+4& 1& 0\\ 16+32& 4+4& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 4\times 2& 1& 0\\ 16(1+2)& 4\times 2& 1\end{array}\right]$ . . . .(i)
and $P^3=\left[\begin{array}{ccc}1& 0& 0\\ 4\times 2& 1& 0\\ 16(1+2)& 4\times 2& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 4\times 3& 1& 0\\ 16(1+2+3)& 4\times 3& 1\end{array}\right]$ . . . (ii)
From symmetry,
$P^{50}=\left[\begin{array}{ccc}1& 0& 0\\ 4\times 50& 1& 0\\ 16(1+2+3+\cdots +50)& 4\times 50& 1\end{array}\right]$
$\because$ $P^{50}$ - Q = I [given]
$\therefore \left[\begin{array}{ccc}1-q_{11}& -q_{12}& -q_{13}\\ 200-q_{21}& 1-q_{22}& -q_{23}\\ 16\times \frac{50}{2}\left(51\right)-q_{31}& 200-q_{32}& 1-q_{33}\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
$\Rightarrow 200-q_{21}=0,\frac{16\times 50\times 51}{2}-q_{31}=0,200-q_{32}=0\therefore q_{21}=200,q_{32}=200,q_{31}=20400Thus,\frac{q_{31}+q_{32}}{q_{21}}=\frac{20400+200}{200}=\frac{20600}{200}=103$