Question

Let $P = \begin{bmatrix}1&0&0\\ 3&1&0\\ 9&3&1\end{bmatrix}$ and $Q = \left[q_{ij}\right]$ be two $3\times3$ 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 = \begin{bmatrix}1&0&0\\ 3&1&0\\ 9&3&1\end{bmatrix} $
$ P^{2} = \begin{bmatrix}1&0&0\\ 3+3 &1&0\\ 9+9+9 &3+3&1\end{bmatrix}$
$ P^{3} = \begin{bmatrix}1&0&0\\ 3+3+3 &1&0\\ 6.9&3+3+3&1\end{bmatrix}$
$ P^{n} = \begin{bmatrix}1&0&0\\ 3n &1&0\\ \frac{n\left(n+1\right)}{2}3^{2} &3n&1\end{bmatrix} $
$ P^{5} = \begin{bmatrix}1&0&0\\ 5.3 &1&0\\ 15.9&5.3&1\end{bmatrix} $
$Q =P^{5} +I_{3} $
$Q = \begin{bmatrix}2&0&0\\ 15 &2&0\\ 135 &15&2\end{bmatrix} $
$ \frac{q_{21} +q_{31}}{q_{32}} = \frac{15+135}{15} = 10 $
Aliter
$ P = \begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}+\begin{pmatrix}0&0&0\\ 3&0&0\\ 9&3&0\end{pmatrix} $
$P=I+X $
$ X =\begin{pmatrix}0&0&0\\ 3&0&0\\ 9&3&0\end{pmatrix} $
$X^{2} =\begin{pmatrix}0&0&0\\ 0&0&0\\ 9&0&0\end{pmatrix} $
$ X^{3} =0$
$ P^{5} =I +5X+10X^{2}$
$ Q = P^{5} +I =2I +5X + 10X^{2}$
$ Q =\begin{pmatrix}2&0&0\\ 0&2&0\\ 0&0&2\end{pmatrix}+\begin{pmatrix}0&0&0\\ 15&0&0\\ 15&15&0\end{pmatrix}+\begin{pmatrix}0&0&0\\ 0&0&0\\ 90&0&0\end{pmatrix} $
$\Rightarrow Q = \begin{pmatrix}2&0&0\\ 15&2&0\\ 135&15&2\end{pmatrix}$