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,

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.