Question

Let \[P=\left[ {{a}_{ij}} \right]\] be a \[3\times 3\] matric and let \[Q=\left( {{b}_{ij}} \right)\] where \[{{b}_{ij}}={{2}^{i+j}}{{a}_{ij}}\] 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

Hint: To solve this question, we need to know the concept of determinant and we should know a few properties of determinant like we can take anything common from any row or column of the determinant. For example,
\[\left| \begin{matrix}
   m{{a}_{1}} & m{{b}_{1}} & m{{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|=m\left| \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|\]
By using these properties, we can find the required answer. In this question, we have to find the determinant of the matrix Q which is equal to \[{{b}_{ij}}\] and \[{{b}_{ij}}={{2}^{i+j}}{{a}_{ij}}\] where \[\left[ {{a}_{ij}} \right]=P\].

Complete step by step solution:
Now, we have been given that \[P=\left[ {{a}_{ij}} \right]\]. So, we can write P as
\[\left| P \right|=\left| \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|\]
And similarly, we can write matrix Q as,
\[\left| Q \right|=\left| \begin{matrix}
   {{b}_{11}} & {{b}_{12}} & {{b}_{13}} \\
   {{b}_{21}} & {{b}_{22}} & {{b}_{23}} \\
   {{b}_{31}} & {{b}_{32}} & {{b}_{33}} \\
\end{matrix} \right|\]
Now, we will put the values of \[{{b}_{ij}}\] using the formula \[{{b}_{ij}}={{2}^{i+j}}{{a}_{ij}}\]. So, we will write the matrix Q as,
\[\left| Q \right|=\left| \begin{matrix}
   {{2}^{1+1}}{{a}_{11}} & {{2}^{1+2}}{{a}_{12}} & {{2}^{1+3}}{{a}_{13}} \\
   {{2}^{2+1}}{{a}_{21}} & {{2}^{2+2}}{{a}_{22}} & {{2}^{2+3}}{{a}_{23}} \\
   {{2}^{3+1}}{{a}_{31}} & {{2}^{3+2}}{{a}_{32}} & {{2}^{3+3}}{{a}_{33}} \\
\end{matrix} \right|\]
We can further write it as,
\[\left| Q \right|=\left| \begin{matrix}
   {{2}^{2}}{{a}_{11}} & {{2}^{3}}{{a}_{12}} & {{2}^{4}}{{a}_{13}} \\
   {{2}^{3}}{{a}_{21}} & {{2}^{4}}{{a}_{22}} & {{2}^{5}}{{a}_{23}} \\
   {{2}^{4}}{{a}_{31}} & {{2}^{5}}{{a}_{32}} & {{2}^{6}}{{a}_{33}} \\
\end{matrix} \right|\]
Now, we know that we can take anything common from any row or column of the determinant. For example, we can take m common from
\[\left| \begin{matrix}
   m{{a}_{1}} & m{{b}_{1}} & m{{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|=m\left| \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|\]
Now, we will take \[{{2}^{2}},{{2}^{3}}\] and \[{{2}^{4}}\] common from first, second, and third-row respectively. So, we can write
\[\left| Q \right|={{2}^{2}}\times {{2}^{3}}\times {{2}^{4}}\left| \begin{matrix}
   {{a}_{11}} & 2{{a}_{12}} & {{2}^{2}}{{a}_{13}} \\
   {{a}_{21}} & 2{{a}_{22}} & {{2}^{2}}{{a}_{23}} \\
   {{a}_{31}} & 2{{a}_{32}} & {{2}^{2}}{{a}_{33}} \\
\end{matrix} \right|\]
\[\left| Q \right|={{2}^{2+3+4}}\left| \begin{matrix}
   {{a}_{11}} & 2{{a}_{12}} & {{2}^{2}}{{a}_{13}} \\
   {{a}_{21}} & 2{{a}_{22}} & {{2}^{2}}{{a}_{23}} \\
   {{a}_{31}} & 2{{a}_{32}} & {{2}^{2}}{{a}_{33}} \\
\end{matrix} \right|\]
\[\left| Q \right|={{2}^{9}}\left| \begin{matrix}
   {{a}_{11}} & 2{{a}_{12}} & {{2}^{2}}{{a}_{13}} \\
   {{a}_{21}} & 2{{a}_{22}} & {{2}^{2}}{{a}_{23}} \\
   {{a}_{31}} & 2{{a}_{32}} & {{2}^{2}}{{a}_{33}} \\
\end{matrix} \right|\]
Now, we will take 2 and \[{{2}^{2}}\] common from the second and third columns respectively. So, we will get,
\[\left| Q \right|={{2}^{9}}\times 2\times {{2}^{2}}\left| \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|\]
And we know that,
\[\left| \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|=\left| P \right|\]
So, we can write
\[\left| Q \right|={{2}^{9+1+2}}\left| P \right|\]
\[\left| Q \right|={{2}^{12}}\left| P \right|\]
Now, we have got the relation between |Q| and |P| and to find the value of |Q| we need to know the value of |P| and in the question, we have been given that |P|= 2. Therefore, we get,
\[\left| Q \right|={{2}^{12}}\times 2\]
\[\left| Q \right|={{2}^{13}}\]
Hence, option (d) is the right answer to this question.

Note: To solve this question, we need to remember that we can express \[\left| \begin{matrix}
   m{{a}_{11}} & m{{a}_{12}} & m{{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|\] as \[m\left| \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|\] , and similarly, we can express \[\left| \begin{matrix}
   n{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   n{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   n{{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|\] as \[n\left| \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|\]. Also, there are possibilities that in a hurry we may forget to put the value of |P| and without putting that we will choose the wrong answer and loose marks.