Let P = begin{bmatrix} 1 & 0 & 0 3 & 1 & 0 | vedclass

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(A) Given $P = \begin{bmatrix} 1 & 0 & 0 \ 3 & 1 & 0 \ 9 & 3 & 1 \end{bmatrix}$.We can write $P = I + A$,where $A = \begin{bmatrix} 0 & 0 & 0 \ 3 &

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$10$

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(A) Given $P = \begin{bmatrix} 1 & 0 & 0 \ 3 & 1 & 0 \ 9 & 3 & 1 \end{bmatrix}$.We can write $P = I + A$,where $A = \begin{bmatrix} 0 & 0 & 0 \ 3 & 0 & 0 \ 9 & 3 & 0 \end{bmatrix}$.Note that $A^2 = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 9 & 0 & 0 \end{bmatrix}$ and $A^3 = O$.Using the binomial expansion for $P^n = (I + A)^n = I + nA + \frac{n(n-1)}{2}A^2$ (since $A^3 = O$):$P^5 = I + 5A + \frac{5 	imes 4}{2}A^2 = I + 5A + 10A^2$.$P^5 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} + 5 \begin{bmatrix} 0 & 0 & 0 \ 3 & 0 & 0 \ 9 & 3 & 0 \end{bmatrix} + 10 \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 9 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 15 & 1 & 0 \ 45+90 & 15 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 15 & 1 & 0 \ 135 & 15 & 1 \end{bmatrix}$.Given $Q = P^5 + I_3 = \begin{bmatrix} 1 & 0 & 0 \ 15 & 1 & 0 \ 135 & 15 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 & 0 \ 15 & 2 & 0 \ 135 & 15 & 2 \end{bmatrix}$.Thus,$q_{21} = 15$,$q_{31} = 135$,and $q_{32} = 15$.The value of $\frac{q_{21} + q_{31}}{q_{32}} = \frac{15 + 135}{15} = \frac{150}{15} = 10$.

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

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