Mathematics - Matrices Question with Solution | TestHub

Given,

$P=\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 9& 3& 1\end{array}\right]$

$\Rightarrow P^2=\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 9& 3& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 9& 3& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 6& 1& 0\\ 27& 6& 1\end{array}\right]$

$\Rightarrow P^4=P^2\times P^2=\left[\begin{array}{ccc}1& 0& 0\\ 6& 1& 0\\ 27& 6& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 6& 1& 0\\ 27& 6& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 12& 1& 0\\ 90& 12& 1\end{array}\right]$

$\Rightarrow P^5=P^4\times P=\left[\begin{array}{ccc}1& 0& 0\\ 12& 1& 0\\ 90& 12& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 9& 3& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 15& 1& 0\\ 135& 15& 1\end{array}\right]$

Let,

$Q=\left[\begin{array}{ccc}{q}_{11}& q_{12}& q_{13}\\ q_{21}& q_{22}& q_{23}\\ q_{31}& q_{31}& q_{33}\end{array}\right]$

Now, given $Q-P^5=I_3$

$\Rightarrow q_{21}-15=0,q_{31}-135=0,q_{32}-15=0$

$\Rightarrow q_{21}=15,q_{31}=135,q_{32}=15$

$\therefore \frac{q_{21}+q_{31}}{q_{32}}=\frac{15+135}{15}=10$.