Question

Let $P=\left[a_{ij}\right]$ be a $3\times 3$ matrix and let $Q=\left[b_{ij}\right]$, where $b_{ij}=2^{i+j}\cdot a_{ij}$ for $1\le i,j\le 3$. 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 1

The correct option is D $2^{13}$

Let,
$P=[a_{ij}{]}_{3\times 3}=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]$

So, $Q=[b_{ij}{]}_{3\times 3}=\left[\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right]$
where, $b_{ij}=2^{i+j}{a}_{ij}$
$\therefore Q=\left[\begin{array}{ccc}4a_{11}& 8a_{12}& 16a_{13}\\ 8a_{21}& 16a_{22}& 32a_{23}\\ 16a_{31}& 32a_{32}& 64a_{33}\end{array}\right]$

Now,
$det\left(Q\right)=\left|\begin{array}{ccc}4a_{11}& 8a_{12}& 16a_{13}\\ 8a_{21}& 16a_{22}& 32a_{23}\\ 16a_{31}& 32a_{32}& 64a_{33}\end{array}\right|$
$=4\times 8\times 16\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ 2a_{21}& 2a_{22}& 2a_{23}\\ 4a_{31}& 4a_{32}& 4a_{33}\end{array}\right|=2^9\times 2\times 4\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|$
$=2^{12}\cdot det\left(P\right)=2^{12}\cdot 2=2^{13}$