Let P=[ … ] and Q=[q_ij] be two 3× 3 matrices such that Q-P^5=I_3. Then q_21+q_31 q_32 is equal to :

Given,

$$P=[\begin{matrix}1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1\end{matrix}]$$P=[139​013​001​]

$$\Rightarrow {P}^{2}=[\begin{matrix}1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1\end{matrix}][\begin{matrix}1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1\end{matrix}]=[\begin{matrix}1 & 0 & 0 \\ 6 & 1 & 0 \\ 27 & 6 & 1\end{matrix}]$$⇒P2=[139​013​001​][139​013​001​]=[1627​016​001​]

$$\Rightarrow {P}^{4}={P}^{2}\times {P}^{2}=[\begin{matrix}1 & 0 & 0 \\ 6 & 1 & 0 \\ 27 & 6 & 1\end{matrix}][\begin{matrix}1 & 0 & 0 \\ 6 & 1 & 0 \\ 27 & 6 & 1\end{matrix}]=[\begin{matrix}1 & 0 & 0 \\ 12 & 1 & 0 \\ 90 & 12 & 1\end{matrix}]$$⇒P4=P2×P2=[1627​016​001​][1627​016​001​]=[11290​0112​001​]

$$\Rightarrow {P}^{5}={P}^{4}\times P=[\begin{matrix}1 & 0 & 0 \\ 12 & 1 & 0 \\ 90 & 12 & 1\end{matrix}][\begin{matrix}1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1\end{matrix}]=[\begin{matrix}1 & 0 & 0 \\ 15 & 1 & 0 \\ 135 & 15 & 1\end{matrix}]$$⇒P5=P4×P=[11290​0112​001​][139​013​001​]=[115135​0115​001​]

Let,

$$Q=[\begin{matrix}{q}_{11} & {q}_{12} & {q}_{13} \\ {q}_{21} & {q}_{22} & {q}_{23} \\ {q}_{31} & {q}_{31} & {q}_{33}\end{matrix}]$$Q=[q11​q21​q31​​q12​q22​q31​​q13​q23​q33​​]

Now, given $Q-{P}^{5}={I}_{3}$Q−P5=I3​

$$\Rightarrow {q}_{21}-15=0, {q}_{31}-135=0, {q}_{32}-15=0$$⇒q21​−15=0,q31​−135=0,q32​−15=0

$$\Rightarrow {q}_{21}=15, {q}_{31}=135, {q}_{32}=15$$⇒q21​=15,q31​=135,q32​=15

$\therefore \frac{{q}_{21}+{q}_{31}}{{q}_{32}}=\frac{15+135}{15}=10$∴q32​q21​+q31​​=1515+135​=10.