
If \[{\rm{X}}\] is a square matrix of order \[{\rm{3}} \times {\rm{3}}\]and \[\lambda \] is a scalar, then find the value of \[adj\left( {\lambda X} \right)\].
A. \[\lambda adj\left( X \right)\]
B. \[{\lambda ^3}adj\left( X \right)\]
C\[{\lambda ^2}adj\left( X \right)\].
D. \[{\lambda ^4}adj\left( X \right)\]
Answer
162k+ views
Hint: In the given question, we need to find the value of \[adj\left( {\lambda X} \right)\]. For this, we have to use the property of adjoint of the matrix such as \[adj\left( {\lambda X} \right) = {\lambda ^{n - 1}}\left( {adj\left( X \right)} \right)\] . After, putting the value of the order of a matrix, we will get the value of \[adj\left( {\lambda X} \right)\].
Formula used: The property of adjoint of matrix \[{\rm{X}}\] is given by
\[\left. {adj\left( {\lambda X} \right)} \right| = {\lambda ^{n - 1}}\left. {\left( {adj\left( X \right)} \right)} \right|\]. Here, \[n\] is the order of the matrix \[{\rm{X}}\].
Complete step by step solution:
We know that \[X\] is a matrix of order \[{\rm{3}} \times {\rm{3}}\].
Also, \[\lambda \] is a scalar factor.
Now, we will find the value of \[adj\left( {\lambda X} \right)\].
By using the property of adjoint of matrix \[{\rm{X}}\] , we get
\[adj\left( {\lambda X} \right) = {\lambda ^{n - 1}}\left( {adj\left( X \right)} \right)\]
Here, \[n = 3\]
Thus, we get \[adj\left( {\lambda X} \right) = {\lambda ^{3 - 1}}\left( {adj\left( X \right)} \right)\]
By simplifying, we get
\[adj\left( {\lambda X} \right) = {\lambda ^2}\left( {adj\left( X \right)} \right)\]
Hence, the value of \[adj\left( {\lambda X} \right)\] is \[{\lambda ^2}\left( {adj\left( X \right)} \right)\].
Therefore, the correct option is (C).
Additional information: The adjoint of the matrix is defined as the transpose of a cofactor matrix of the square matrix. The adjoint of the matrix \[A\] is denoted by \[adj\left( A \right)\]. It is essential to determine the adjoint of a matrix to calculate the inverse of a matrix. This can be applied only to square matrices.
Note: Students may make mistakes while applying the property of adjoint of matrix. We can solve this example, in another way which is shown below.
Let \[A\]be any \[{\rm{3}} \times {\rm{3}}\] matrix.
We know that, \[{A^{ - 1}} = \dfrac{{adj\left( A \right)}}{{\left| A \right|}}\]
So, \[adj\left( A \right) = {A^{ - 1}}\left| A \right|\]
Now, we can say that \[adj\left( {\lambda A} \right) = {\left( {\lambda A} \right)^{ - 1}}\left| {\lambda A} \right|\]
Here, \[\lambda A\] is a scaling factor.
So, we get \[adj\left( {\lambda A} \right) = {\lambda ^{ - 1}}{\left( A \right)^{ - 1}}\left| {\lambda A} \right|\]
This gives, \[adj\left( {\lambda A} \right) = \dfrac{1}{\lambda }{\left( A \right)^{ - 1}}\left| {\lambda A} \right|\]
But we know that \[\left| {kA} \right| = {k^n}\left| A \right|\], \[n\]is the order of the matrix \[A\].
Thus, we get \[adj\left( {\lambda A} \right) = \dfrac{1}{\lambda }{\left( A \right)^{ - 1}}\left( {{\lambda ^n}} \right)\left| A \right|\]
Here, \[n = 3\]
Hence, we get \[adj\left( {\lambda A} \right) = \dfrac{1}{\lambda }{\left( A \right)^{ - 1}}\left( {{\lambda ^3}} \right)\left| A \right|\]
By simplifying, we get \[adj\left( {\lambda A} \right) = {\left( A \right)^{ - 1}}\left( {{\lambda ^2}} \right)\left| A \right|\]
Therefore, \[adj\left( {\lambda A} \right) = \left( {{\lambda ^2}} \right){\left( A \right)^{ - 1}}\left| A \right|\]
But \[adj\left( A \right) = {A^{ - 1}}\left| A \right|\]
So, we get \[adj\left( {\lambda A} \right) = \left( {{\lambda ^2}} \right)adj\left( A \right)\]
Therefore, the result of both methods is the same.
Formula used: The property of adjoint of matrix \[{\rm{X}}\] is given by
\[\left. {adj\left( {\lambda X} \right)} \right| = {\lambda ^{n - 1}}\left. {\left( {adj\left( X \right)} \right)} \right|\]. Here, \[n\] is the order of the matrix \[{\rm{X}}\].
Complete step by step solution:
We know that \[X\] is a matrix of order \[{\rm{3}} \times {\rm{3}}\].
Also, \[\lambda \] is a scalar factor.
Now, we will find the value of \[adj\left( {\lambda X} \right)\].
By using the property of adjoint of matrix \[{\rm{X}}\] , we get
\[adj\left( {\lambda X} \right) = {\lambda ^{n - 1}}\left( {adj\left( X \right)} \right)\]
Here, \[n = 3\]
Thus, we get \[adj\left( {\lambda X} \right) = {\lambda ^{3 - 1}}\left( {adj\left( X \right)} \right)\]
By simplifying, we get
\[adj\left( {\lambda X} \right) = {\lambda ^2}\left( {adj\left( X \right)} \right)\]
Hence, the value of \[adj\left( {\lambda X} \right)\] is \[{\lambda ^2}\left( {adj\left( X \right)} \right)\].
Therefore, the correct option is (C).
Additional information: The adjoint of the matrix is defined as the transpose of a cofactor matrix of the square matrix. The adjoint of the matrix \[A\] is denoted by \[adj\left( A \right)\]. It is essential to determine the adjoint of a matrix to calculate the inverse of a matrix. This can be applied only to square matrices.
Note: Students may make mistakes while applying the property of adjoint of matrix. We can solve this example, in another way which is shown below.
Let \[A\]be any \[{\rm{3}} \times {\rm{3}}\] matrix.
We know that, \[{A^{ - 1}} = \dfrac{{adj\left( A \right)}}{{\left| A \right|}}\]
So, \[adj\left( A \right) = {A^{ - 1}}\left| A \right|\]
Now, we can say that \[adj\left( {\lambda A} \right) = {\left( {\lambda A} \right)^{ - 1}}\left| {\lambda A} \right|\]
Here, \[\lambda A\] is a scaling factor.
So, we get \[adj\left( {\lambda A} \right) = {\lambda ^{ - 1}}{\left( A \right)^{ - 1}}\left| {\lambda A} \right|\]
This gives, \[adj\left( {\lambda A} \right) = \dfrac{1}{\lambda }{\left( A \right)^{ - 1}}\left| {\lambda A} \right|\]
But we know that \[\left| {kA} \right| = {k^n}\left| A \right|\], \[n\]is the order of the matrix \[A\].
Thus, we get \[adj\left( {\lambda A} \right) = \dfrac{1}{\lambda }{\left( A \right)^{ - 1}}\left( {{\lambda ^n}} \right)\left| A \right|\]
Here, \[n = 3\]
Hence, we get \[adj\left( {\lambda A} \right) = \dfrac{1}{\lambda }{\left( A \right)^{ - 1}}\left( {{\lambda ^3}} \right)\left| A \right|\]
By simplifying, we get \[adj\left( {\lambda A} \right) = {\left( A \right)^{ - 1}}\left( {{\lambda ^2}} \right)\left| A \right|\]
Therefore, \[adj\left( {\lambda A} \right) = \left( {{\lambda ^2}} \right){\left( A \right)^{ - 1}}\left| A \right|\]
But \[adj\left( A \right) = {A^{ - 1}}\left| A \right|\]
So, we get \[adj\left( {\lambda A} \right) = \left( {{\lambda ^2}} \right)adj\left( A \right)\]
Therefore, the result of both methods is the same.
Recently Updated Pages
JEE Advanced Course 2025 - Subject List, Syllabus, Course, Details

Crack JEE Advanced 2025 with Vedantu's Live Classes

JEE Advanced Maths Revision Notes

JEE Advanced Chemistry Revision Notes

Download Free JEE Advanced Revision Notes PDF Online for 2025

Solutions Class 12 Notes JEE Advanced Chemistry [PDF]

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

IIT Fees Structure 2025

Top IIT Colleges in India 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

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

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

Degree of Dissociation and Its Formula With Solved Example for JEE
