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# Let $P = \left[ {{a_{ij}}} \right]$ be a $3 \times 3$ matrix and let $Q = \left[ {{b_{ij}}} \right]$, where ${b_{ij}} = {2^{i + j}}{a_{ij}}$ , for $1 \leqslant i,j \leqslant 3$.If the determinant of $P$ is $2$,then the determinant of the matrix $Q$ is${\text{A}}{\text{. }}{{\text{2}}^{10}} \\ {\text{B}}{\text{. }}{{\text{2}}^{11}} \\ {\text{C}}{\text{. }}{{\text{2}}^{12}} \\ {\text{D}}{\text{. }}{{\text{2}}^{13}} \\$ Verified
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Hint: In this question we will have to understand the concept of matrices and their determinants and the relation between them and then analyse the question and given options one by one to find the correct answer.

According to question
${\text{Q = }}\left[ {{{\text{b}}_{ij}}} \right]$ where ${{\text{b}}_{ij}} = {2^{i + j}}{a_{ij}}$
Therefore the determinant of ${\text{Q}}$
$\left| {\text{Q}} \right| = \left| {\begin{array}{*{20}{c}} {{2^2}{a_{11}}}&{{2^3}{a_{12}}}&{{2^4}{a_{13}}} \\ {{2^3}{a_{21}}}&{{2^4}{a_{22}}}&{{2^5}{a_{23}}} \\ {{2^4}{a_{31}}}&{{2^5}{a_{32}}}&{{2^6}{a_{33}}} \end{array}} \right|$
From first row we’ll take ${2^2}$ common then from ${2^{nd}}$ row take ${2^3}$ common whereas from ${3^{rd}}$ row take ${2^4}$ common. From this we get
$\left| {\text{Q}} \right| = {2^2} \times {2^3} \times {2^4}\left| {\begin{array}{*{20}{c}} {{{\text{a}}_{11}}}&{2{{\text{a}}_{12}}}&{{2^2}{{\text{a}}_{13}}} \\ {{{\text{a}}_{21}}}&{2{{\text{a}}_{22}}}&{{2^2}{{\text{a}}_{23}}} \\ {{{\text{a}}_{31}}}&{2{{\text{a}}_{32}}}&{{2^2}{{\text{a}}_{33}}} \end{array}} \right|$
Again from ${2^{{\text{nd}}}}$ column $2$ is taken common and from ${3^{{\text{rd}}}}$ column ${2^2}$ is taken common we get,
$\left| {\text{Q}} \right| = {2^9} \times 2 \times {2^2}\left| {\begin{array}{*{20}{c}} {{{\text{a}}_{11}}}&{{{\text{a}}_{12}}}&{{{\text{a}}_{13}}} \\ {{{\text{a}}_{21}}}&{{{\text{a}}_{22}}}&{{{\text{a}}_{23}}} \\ {{{\text{a}}_{31}}}&{{{\text{a}}_{32}}}&{{{\text{a}}_{33}}} \end{array}} \right|$
Now, according to question determinant of ${\text{P}}$ is $2$ and ${\text{P = }}\left[ {{{\text{a}}_{ij}}} \right]$
$\therefore \left| {\text{Q}} \right| = {2^{12}}\left| {\text{P}} \right| = {2^{12}} \times 2 = {2^{13}}$

Note: To solve such types of questions we have to understand that determinant is a scalar value that can be computed from the elements of a square matrix. Determinant of a matrix ${\text{A}}$is denoted by $\det \left( {\text{A}} \right),\left| {\text{A}} \right|$ or $\det$A .
Last updated date: 28th Sep 2023
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