Let $$P = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{bmatrix}$$ and $$Q = [q_{ij}]$$ be two $$3 \times 3$$ matrices such that $$Q - P^5 = I_3$$. Then $$\frac{q_{21} + q_{31}}{q_{32}}$$ is equal to:

We have the two given matrices

$$P=\begin{bmatrix}1&0&0\\3&1&0\\9&3&1\end{bmatrix}\qquad\text{and}\qquad Q=[q_{ij}]$$

together with the relation

$$Q-P^{5}=I_{3}.$$

This implies immediately that

$$Q=P^{5}+I_{3}.$$

So our task is to find the fifth power of the matrix $$P$$, add the identity matrix $$I_{3}$$ to it, and then read off the required entries of $$Q$$.

Because $$P$$ is a lower-triangular matrix whose diagonal entries are all $$1$$, every power of $$P$$ will also be lower-triangular with diagonal entries $$1$$. We proceed by repeated multiplication, writing out every step clearly.

First power (already known):

$$P^{1}=P=\begin{bmatrix}1&0&0\\3&1&0\\9&3&1\end{bmatrix}.$$

Second power: we use the definition of matrix multiplication. For the entry in the second row and first column, for example, we multiply the second row of the left matrix by the first column of the right matrix. Carrying this out for every position we get

$$\begin{aligned} P^{2}&=P\cdot P\\ &=\begin{bmatrix} 1\cdot1+0\cdot3+0\cdot9 & 1\cdot0+0\cdot1+0\cdot3 & 1\cdot0+0\cdot0+0\cdot1\\ 3\cdot1+1\cdot3+0\cdot9 & 3\cdot0+1\cdot1+0\cdot3 & 3\cdot0+1\cdot0+0\cdot1\\ 9\cdot1+3\cdot3+1\cdot9 & 9\cdot0+3\cdot1+1\cdot3 & 9\cdot0+3\cdot0+1\cdot1 \end{bmatrix}\\[6pt] &=\begin{bmatrix} 1&0&0\\ 6&1&0\\ 27&6&1 \end{bmatrix}. \end{aligned}$$

Third power: multiply $$P^{2}$$ by $$P$$.

$$\begin{aligned} P^{3}&=P^{2}\cdot P\\ &=\begin{bmatrix} 1&0&0\\ 6&1&0\\ 27&6&1 \end{bmatrix} \begin{bmatrix} 1&0&0\\ 3&1&0\\ 9&3&1 \end{bmatrix}\\ &=\begin{bmatrix} 1\cdot1+0\cdot3+0\cdot9 & 1\cdot0+0\cdot1+0\cdot3 & 1\cdot0+0\cdot0+0\cdot1\\[4pt] 6\cdot1+1\cdot3+0\cdot9 & 6\cdot0+1\cdot1+0\cdot3 & 6\cdot0+1\cdot0+0\cdot1\\[4pt] 27\cdot1+6\cdot3+1\cdot9 & 27\cdot0+6\cdot1+1\cdot3 & 27\cdot0+6\cdot0+1\cdot1 \end{bmatrix}\\[6pt] &=\begin{bmatrix} 1&0&0\\ 9&1&0\\ 54&9&1 \end{bmatrix}. \end{aligned}$$

Fourth power: multiply $$P^{3}$$ by $$P$$ once more.

$$\begin{aligned} P^{4}&=P^{3}\cdot P\\ &=\begin{bmatrix} 1&0&0\\ 9&1&0\\ 54&9&1 \end{bmatrix} \begin{bmatrix} 1&0&0\\ 3&1&0\\ 9&3&1 \end{bmatrix}\\ &=\begin{bmatrix} 1&0&0\\ 9\cdot1+1\cdot3+0\cdot9 & 9\cdot0+1\cdot1+0\cdot3 & 0\\ 54\cdot1+9\cdot3+1\cdot9 & 54\cdot0+9\cdot1+1\cdot3 & 1 \end{bmatrix}\\[6pt] &=\begin{bmatrix} 1&0&0\\ 12&1&0\\ 90&12&1 \end{bmatrix}. \end{aligned}$$

Fifth power: one final multiplication of $$P^{4}$$ by $$P$$.

$$\begin{aligned} P^{5}&=P^{4}\cdot P\\ &=\begin{bmatrix} 1&0&0\\ 12&1&0\\ 90&12&1 \end{bmatrix} \begin{bmatrix} 1&0&0\\ 3&1&0\\ 9&3&1 \end{bmatrix}\\ &=\begin{bmatrix} 1&0&0\\ 12\cdot1+1\cdot3+0\cdot9 & 12\cdot0+1\cdot1+0\cdot3 & 0\\ 90\cdot1+12\cdot3+1\cdot9 & 90\cdot0+12\cdot1+1\cdot3 & 1 \end{bmatrix}\\[6pt] &=\begin{bmatrix} 1&0&0\\ 15&1&0\\ 135&15&1 \end{bmatrix}. \end{aligned}$$

Now we add the identity matrix $$I_{3}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$$ to obtain $$Q$$:

$$Q=P^{5}+I_{3} =\begin{bmatrix} 1+1 & 0 & 0\\ 15 & 1+1 & 0\\ 135 & 15 & 1+1 \end{bmatrix} =\begin{bmatrix} 2&0&0\\ 15&2&0\\ 135&15&2 \end{bmatrix}.$$

From this explicit form we read off

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

We are asked to compute

$$\frac{q_{21}+q_{31}}{q_{32}}=\frac{15+135}{15}=\frac{150}{15}=10.$$

Hence, the correct answer is Option A.