Prove the following
Question:

Let $P=\left[\begin{array}{lll}1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1\end{array}\right]$ and $\mathrm{Q}=\left[\mathrm{q}_{\mathrm{ij}}\right]$ be two $3 \times 3$

matrices such that $\mathrm{Q}-\mathrm{P}^{5}=\mathrm{I}_{3}$. Then $\frac{\mathrm{q}_{21}+\mathrm{q}_{31}}{\mathrm{q}_{32}}$ is qual to :

1. 15

2. 9

3. 135

4. 10

Correct Option: , 4

Solution:

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

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

$\mathrm{P}^{3}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 3+3+3 & 1 & 0 \\ 6.9 & 3+3+3 & 1\end{array}\right]$

$P^{n}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 n & 1 & 0 \\ \frac{n(n+1)}{2} 3^{2} & 3 n & 1\end{array}\right]$

$\mathrm{P}^{5}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 5.3 & 1 & 0 \\ 15.9 & 5.3 & 1\end{array}\right]$

$\mathrm{Q}=\mathrm{P}^{5}+\mathrm{I}_{3}$

$\mathrm{Q}=\left[\begin{array}{ccc}2 & 0 & 0 \\ 15 & 2 & 0 \\ 135 & 15 & 2\end{array}\right]$

$\frac{\mathrm{q}_{21}+\mathrm{q}_{31}}{\mathrm{q}_{32}}=\frac{15+135}{15}=10$

Aliter

$P=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)+\left(\begin{array}{lll}0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0\end{array}\right)$

$P=I+X$

$X=\left(\begin{array}{lll}0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0\end{array}\right)$

$\mathbf{X}^{2}=\left(\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0\end{array}\right)$

$\mathrm{X}^{3}=0$

$\mathrm{P}^{5}=\mathrm{I}+5 \mathrm{X}+10 \mathrm{X}^{2}$

$Q=P^{5}+I=2 I+5 X+10 X^{2}$

$Q=\left(\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right)+\left(\begin{array}{ccc}0 & 0 & 0 \\ 15 & 0 & 0 \\ 15 & 15 & 0\end{array}\right)+\left(\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 90 & 0 & 0\end{array}\right)$

$\Rightarrow Q=\left(\begin{array}{ccc}2 & 0 & 0 \\ 15 & 2 & 0 \\ 135 & 15 & 2\end{array}\right)$