Question

Let $P=\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 9& 3& 1\end{array}\right]$ and $Q=\left[q_{ij}\right]$ be two $3\times 3$ matrices such that $Q-p^5=I_3$. Then $\frac{q_{21}+q_{31}}{q32}$ is equal to

• A. 15
• B. 9
• C. 135
• D. 10

Answer 1

Answer: (D)

$P=\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 9& 3& 1\end{array}\right]$
$P^2=\left[\begin{array}{ccc}1& 0& 0\\ 3+3& 1& 0\\ 9+9+9& 3+3& 1\end{array}\right]$
$P^3=\left[\begin{array}{ccc}1& 0& 0\\ 3+3+3& 1& 0\\ 6.9& 3+3+3& 1\end{array}\right]$
$P^n=\left[\begin{array}{ccc}1& 0& 0\\ 3n& 1& 0\\ \frac{n\left(n+1\right)}{2}{3}^2& 3n& 1\end{array}\right]$
$P^5=\left[\begin{array}{ccc}1& 0& 0\\ 5.3& 1& 0\\ 15.9& 5.3& 1\end{array}\right]$
$Q=P^5+I_3$
$Q=\left[\begin{array}{ccc}2& 0& 0\\ 15& 2& 0\\ 135& 15& 2\end{array}\right]$
$\frac{q_{21}+q_{31}}{q_{32}}=\frac{15+135}{15}=10$
Aliter
$P=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)+\left(\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 9& 3& 0\end{array}\right)$
$P=I+X$
$X=\left(\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 9& 3& 0\end{array}\right)$
$X^2=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 9& 0& 0\end{array}\right)$
$X^3=0$
$P^5=I+5X+10X^2$
$Q=P^5+I=2I+5X+10X^2$
$Q=\left(\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right)+\left(\begin{array}{ccc}0& 0& 0\\ 15& 0& 0\\ 15& 15& 0\end{array}\right)+\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 90& 0& 0\end{array}\right)$
$\Rightarrow Q=\left(\begin{array}{ccc}2& 0& 0\\ 15& 2& 0\\ 135& 15& 2\end{array}\right)$