
If \[A = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&1\\0&2&{ - 3}\\2&1&0\end{array}} \right]\] , \[B = \left( {adj A} \right)\], and \[C = 5A\]. Then find the value of \[\dfrac{{\left| {adj B} \right|}}{{\left| C \right|}}\].
A. 5
B. \[25\]
C. \[ - 1\]
D. 1
E. \[125\]
Answer
162.6k+ views
Hint: First, calculate the determinant of the given \[3 \times 3\] matrix \[A\]. Simplify the required expression by using the given equations. Then apply the properties of the determinant of the matrix and solve it to get the required answer.
Formula used:
The determinant of a \[3 \times 3\] matrix \[A = \left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right]\] is: \[\left| A \right| = {a_{11}}\left( {{a_{22}}{a_{33}} - {a_{32}}{a_{23}}} \right) - {a_{12}}\left( {{a_{21}}{a_{33}} - {a_{31}}{a_{23}}} \right) + {a_{13}}\left( {{a_{21}}{a_{32}} - {a_{31}}{a_{22}}} \right)\]
The determinant of a scalar \[m\] times a matrix \[A\] of order \[n\] is: \[\left| {mA} \right| = {m^n}\left| A \right|\]
\[adj\left( {adj A} \right) = {\left| A \right|^{n - 2}}A\]
\[\left| {adj\left( {adj A} \right)} \right| = \left| {{{\left| A \right|}^{n - 2}}A} \right| = {\left( {{{\left| A \right|}^{n - 2}}} \right)^n}\left| A \right|\]
Complete step by step solution:
The given matrices are \[A = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&1\\0&2&{ - 3}\\2&1&0\end{array}} \right]\], \[B = \left( {adj A} \right)\], and \[C = 5A\].
Let’s calculate the determinant of the given matrix \[A\].
Apply the formula of the determinant of a \[3 \times 3\] matrix \[\left| A \right| = {a_{11}}\left( {{a_{22}}{a_{33}} - {a_{32}}{a_{23}}} \right) - {a_{12}}\left( {{a_{21}}{a_{33}} - {a_{31}}{a_{23}}} \right) + {a_{13}}\left( {{a_{21}}{a_{32}} - {a_{31}}{a_{23}}} \right)\].
We get,
\[\left| A \right| = 1\left( {2 \times 0 - 1 \times \left( { - 3} \right)} \right) - \left( { - 1} \right)\left( {0 \times 0 - 2 \times \left( { - 3} \right)} \right) + 1\left( {0 \times 1 - 2 \times 2} \right)\]
\[ \Rightarrow \left| A \right| = 1\left( 3 \right) + 1\left( 6 \right) + 1\left( { - 4} \right)\]
\[ \Rightarrow \left| A \right| = 3 + 6 - 4\]
\[ \Rightarrow \left| A \right| = 5\]
Now solve the required expression \[\dfrac{{\left| {adj B} \right|}}{{\left| C \right|}}\] by using the given equations \[B = \left( {adj A} \right)\], and \[C = 5A\].
\[\dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{\left| {adj \left( {adj A} \right)} \right|}}{{\left| {5A} \right|}}\]
Use the properties for the determinant.
\[\dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{\left| {{{\left| A \right|}^{n - 2}}A} \right|}}{{{5^n}\left| A \right|}}\]
\[ \Rightarrow \dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{{{\left( {{{\left| A \right|}^{n - 2}}} \right)}^n}\left| A \right|}}{{{5^n}\left| A \right|}}\]
Here, the order of the square matrix is 3.
Then,
\[\dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{{{\left( {{{\left| A \right|}^{3 - 2}}} \right)}^3}\left| A \right|}}{{{5^3}\left| A \right|}}\]
\[ \Rightarrow \dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{{{\left( {{{\left| A \right|}^1}} \right)}^3}\left| A \right|}}{{{5^3}\left| A \right|}}\]
\[ \Rightarrow \dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{{{\left| A \right|}^3}\left| A \right|}}{{{5^3}\left| A \right|}}\]
Now substitute the value of the determinant in the above equation.
\[\dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{{5^3} \times 5}}{{{5^3} \times 5}}\]
Cancel out the common terms from the numerator and the denominator.
\[\dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = 1\]
Hence the correct option is D.
Note: Students should keep in mind that the adjoint matrix of any matrix is the transpose of its cofactor matrix. While solving the questions related to the properties of the adjoint matrix, students can make note of the following important adjoint properties:
\[A\left( {adj A} \right) = \left( {adj A} \right)A = \left| A \right|I\]
\[\left| {adj A} \right| = {\left| A \right|^{n - 1}}\], where \[n\] is the order of the square matrix
\[adj\left( {adj A} \right) = {\left| A \right|^{n - 2}}A\]
\[adj\left( {AB} \right) = adj\left( B \right)adj\left( A \right)\]
Formula used:
The determinant of a \[3 \times 3\] matrix \[A = \left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right]\] is: \[\left| A \right| = {a_{11}}\left( {{a_{22}}{a_{33}} - {a_{32}}{a_{23}}} \right) - {a_{12}}\left( {{a_{21}}{a_{33}} - {a_{31}}{a_{23}}} \right) + {a_{13}}\left( {{a_{21}}{a_{32}} - {a_{31}}{a_{22}}} \right)\]
The determinant of a scalar \[m\] times a matrix \[A\] of order \[n\] is: \[\left| {mA} \right| = {m^n}\left| A \right|\]
\[adj\left( {adj A} \right) = {\left| A \right|^{n - 2}}A\]
\[\left| {adj\left( {adj A} \right)} \right| = \left| {{{\left| A \right|}^{n - 2}}A} \right| = {\left( {{{\left| A \right|}^{n - 2}}} \right)^n}\left| A \right|\]
Complete step by step solution:
The given matrices are \[A = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&1\\0&2&{ - 3}\\2&1&0\end{array}} \right]\], \[B = \left( {adj A} \right)\], and \[C = 5A\].
Let’s calculate the determinant of the given matrix \[A\].
Apply the formula of the determinant of a \[3 \times 3\] matrix \[\left| A \right| = {a_{11}}\left( {{a_{22}}{a_{33}} - {a_{32}}{a_{23}}} \right) - {a_{12}}\left( {{a_{21}}{a_{33}} - {a_{31}}{a_{23}}} \right) + {a_{13}}\left( {{a_{21}}{a_{32}} - {a_{31}}{a_{23}}} \right)\].
We get,
\[\left| A \right| = 1\left( {2 \times 0 - 1 \times \left( { - 3} \right)} \right) - \left( { - 1} \right)\left( {0 \times 0 - 2 \times \left( { - 3} \right)} \right) + 1\left( {0 \times 1 - 2 \times 2} \right)\]
\[ \Rightarrow \left| A \right| = 1\left( 3 \right) + 1\left( 6 \right) + 1\left( { - 4} \right)\]
\[ \Rightarrow \left| A \right| = 3 + 6 - 4\]
\[ \Rightarrow \left| A \right| = 5\]
Now solve the required expression \[\dfrac{{\left| {adj B} \right|}}{{\left| C \right|}}\] by using the given equations \[B = \left( {adj A} \right)\], and \[C = 5A\].
\[\dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{\left| {adj \left( {adj A} \right)} \right|}}{{\left| {5A} \right|}}\]
Use the properties for the determinant.
\[\dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{\left| {{{\left| A \right|}^{n - 2}}A} \right|}}{{{5^n}\left| A \right|}}\]
\[ \Rightarrow \dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{{{\left( {{{\left| A \right|}^{n - 2}}} \right)}^n}\left| A \right|}}{{{5^n}\left| A \right|}}\]
Here, the order of the square matrix is 3.
Then,
\[\dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{{{\left( {{{\left| A \right|}^{3 - 2}}} \right)}^3}\left| A \right|}}{{{5^3}\left| A \right|}}\]
\[ \Rightarrow \dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{{{\left( {{{\left| A \right|}^1}} \right)}^3}\left| A \right|}}{{{5^3}\left| A \right|}}\]
\[ \Rightarrow \dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{{{\left| A \right|}^3}\left| A \right|}}{{{5^3}\left| A \right|}}\]
Now substitute the value of the determinant in the above equation.
\[\dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{{5^3} \times 5}}{{{5^3} \times 5}}\]
Cancel out the common terms from the numerator and the denominator.
\[\dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = 1\]
Hence the correct option is D.
Note: Students should keep in mind that the adjoint matrix of any matrix is the transpose of its cofactor matrix. While solving the questions related to the properties of the adjoint matrix, students can make note of the following important adjoint properties:
\[A\left( {adj A} \right) = \left( {adj A} \right)A = \left| A \right|I\]
\[\left| {adj A} \right| = {\left| A \right|^{n - 1}}\], where \[n\] is the order of the square matrix
\[adj\left( {adj A} \right) = {\left| A \right|^{n - 2}}A\]
\[adj\left( {AB} \right) = adj\left( B \right)adj\left( A \right)\]
Recently Updated Pages
JEE Advanced 2022 Maths Question Paper 2 with Solutions

JEE Advanced Study Plan 2025: Expert Tips and Preparation Guide

JEE Advanced 2022 Physics Question Paper 2 with Solutions

Carbohydrates Class 12 Important Questions JEE Advanced Chemistry [PDF]

JEE Advanced 2022 Question Paper with Solutions PDF free Download

JEE Advanced 2025 Surface Chemistry Revision Notes - Free PDF Download

Trending doubts
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

IIT CSE Cutoff: Category‐Wise Opening and Closing Ranks

Top IIT Colleges in India 2025

IIT Fees Structure 2025

IIT Roorkee Average Package 2025: Latest Placement Trends Updates

Other Pages
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More
