Question
Let $$P = \left[ {{a_{ij}}} \right]{\text{be a 3}} \times {\text{3}}$$ matrix and let $$Q = \left[ {{b_{ij}}} \right],{\text{where }}{b_{ij}} = {2^{i + j}}{a_{ij}}\,{\text{for 1}} \leqslant i,j \leqslant 3.$$ If the determinant of $$P$$ is 2, then the determinant of the matrix $$Q$$ is
A.
$${2^{10}}$$
B.
$${2^{11}}$$
C.
$${2^{12}}$$
D.
$${2^{13}}$$
Answer :
$${2^{13}}$$
Solution :
We have
\[\begin{array}{l}
\left| Q \right| = \left| \begin{array}{l}
{2^2}{a_{11}}\,\,\,\,\,\,\,\,{2^3}{a_{12}}\,\,\,\,\,\,\,\,\,{2^4}{a_{13}}\\
{2^3}{a_{21}}\,\,\,\,\,\,\,\,{2^4}{a_{22}}\,\,\,\,\,\,\,\,{2^5}{a_{23}}\\
{2^4}{a_{31}}\,\,\,\,\,\,\,\,{2^5}{a_{32}}\,\,\,\,\,\,\,\,{2^6}{a_{33}}
\end{array} \right|\\
= {2^2}{.2^3}{.2^4}\,\left| \begin{array}{l}
\,\,\,{a_{11}}\,\,\,\,\,\,\,\,\,\,{a_{12}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{a_{13}}\\
\,2{a_{21}}\,\,\,\,\,\,\,2{a_{22}}\,\,\,\,\,\,\,\,\,\,\,\,2{a_{23}}\\
{2^2}{a_{31}}\,\,\,\,\,{2^2}{a_{32}}\,\,\,\,\,\,\,\,\,{2^2}{a_{33}}
\end{array} \right|\\
= {2^9}{.2.2^2}\left| \begin{array}{l}
{a_{11}}\,\,\,\,\,{a_{12}}\,\,\,\,\,\,{a_{13}}\\
{a_{21}}\,\,\,\,\,{a_{22}}\,\,\,\,\,{a_{23}}\\
{a_{31}}\,\,\,\,\,{a_{32}}\,\,\,\,\,{a_{33}}
\end{array} \right|
\end{array}\]
$$ = {2^{12}} \times \left| P \right| = {2^{12}} \times 2 = {2^{13}}$$